The PiXL Club The PiXL Club The PiXL

The PiXL Club The PiXL Club The PiXL

The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club XL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The

PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The

PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club Theresource PiXL Club The PiXL Club PiXL Club PiXL Club The PiXLfor Club PiXLas Club Theremain PiXL Clubmembers

The PiXL Club The PiXL Club The PiXL Clubnot Thebe PiXLcopied, Club Thesold PiXL nor Clubtransferred The PiXL Clubto The This is strictly forThe the use

ofThe member schools asThe long they of The PiXL Club. It may PiXL Club The PiXL Club The PiXL Club

The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL

Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The

a third party or used by the school after membership ceases. Until such time it may be freely used within the member school. PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The All opinions and contributions are those of the authors. The contents of this resource are not connected with nor endorsed by any other company, PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The organisation institution. PiXL Club The PiXL Clubor The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The www.pixl.org.uk The PiXL Club Ltd, Company number 07321607 PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The

PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL Club The PiXL KnowIT! GCSE Chemistry AQA Topic Quantitative chemistry Copyright The PiXL Club Ltd, 2017 Overview Chemical measurements Balanced chemical equations Conservation of mass Relative formula mass Use of amount of substance (HT) Amounts of substances in equations (HT) Quantities in equations (HT) Using moles to balance equations (HT)

Limiting reactants (HT) Concentrations of solutions Quantities (chemistry only) Percentage yield Atom economy Moles of solutions and gases (HT) Quantitative chemistry LearnIT! KnowIT! Chemical Measurements PART 1 Balanced chemical equations Conservation of mass

Chemical measurements - PART 1 Chemical equations can be very useful. The law of conservation states that no atoms are lost or made during a chemical reaction so the mass of the product equals the mass of the reactants. Chemical reactions can be represented by symbol equations which are balanced in terms of the numbers of atoms of each element involved on both sides of the equation. State symbols s, l, g and aq are used in symbol equations. When hydrogen molecules react with chlorine molecules, they make hydrogen chloride molecules: H2 + Cl2 HCl This equation shows the reactants and products, but it is not balanced. H2 + Cl2 2HCl This balanced equation shows that one hydrogen molecules reacts with one chlorine molecule to form two molecules of hydrochloric acid. Chemical measurements - PART 1

When magnesium is heated in a crucible it reacts with oxygen and forms magnesium oxide: 2Mg + O2 2MgO This equation shows that two magnesium atoms react with one oxygen molecule to form two magnesium oxide compounds. Here are the results from the reaction: Mass in g Mass of crucible at the start of the reaction 0.23 Mass of crucible at end of reaction

0.41 Some reactions may appear to involve a change in mass but this can be explained because a reactant or product is usually a gas and its mass has not been taken into account. In this example, the mass of the magnesium oxide produced is greater than the mass of the original metal. Chemical measurements - PART 1 When calcium carbonate thermally decomposes it forms calcium oxide and carbon dioxide: CaCO3 CaO + CO2 This equation shows that one calcium carbonate

compound (made from one calcium, one carbon and three oxygen atoms) forms one calcium oxide compound and one carbon dioxide molecule. In thermal decomposition of metal carbonates, carbon dioxide is produced Here are the results from the reaction: and escapes into the atmosphere Mass in g leaving the metal oxide as the only Mass of metal carbonate at the 0.54 solid product. start of the reaction In this example, the mass of the Mass of metal carbonate at 0.36

calcium oxide produced is less than the end of reaction mass of the metal carbonate formed. Whenever a measurement is taken, there is always some uncertainty about the result obtained that may have come from a variety of sources within the investigation. It is useful to determine whether the mean value falls within the range of uncertainty of the result. Chemical measurements - PART 1 When calcium carbonate thermally decomposes it forms calcium oxide and carbon dioxide: CaCO3 CaO + CO2 This equation shows that one calcium carbonate compound (made from one calcium, one carbon and three oxygen atoms) forms one calcium oxide compound and one carbon dioxide molecule.

In thermal decomposition of metal carbonates, carbon dioxide is produced Here are the results from the reaction: and escapes into the atmosphere Mass in g leaving the metal oxide as the only Mass of metal carbonate at the 0.54 solid product. start of the reaction In this example, the mass of the Mass of metal carbonate at 0.36 calcium oxide produced is less than the end of reaction mass of the metal carbonate formed. Whenever a measurement is taken, there is always some uncertainty about the result obtained

that may have come from a variety of sources within the investigation. It is useful to determine whether the mean value falls within the range of uncertainty of the result. QuestionIT! Chemical Measurements PART 1 Balanced chemical equations Conservation of mass Chemical measurements QuestionIT 1. What is the law of conservation of mass? 2. Why might some reactions appear to show a change in mass? 3. Give two examples of a reaction where a change in mass may appear to take place. 4. Balance the following equations: a) H2 + O2 H2O

b) Ca + HCl CaCl2 + H2 c) Li + H2O LiOH + H2 d) NH3 + O2 NO + H2O e) K + O2 K2O Chemical measurements QuestionIT 5. How many atoms and elements are in the compound sodium aluminate, NaAl(OH)4? 6. What do the following formulae tell you? a) 2HCl b) Cl2 7. An aqueous solution of hydrogen peroxide (H2O2) decomposes to form water and oxygen. a) Write a balanced symbol equation for this reaction. Include the state symbols. b) Why does the water, produced during the reaction, have a

lower mass than the original hydrogen peroxide? AnswerIT! Chemical Measurements PART 1 Balanced chemical equations Conservation of mass Chemical measurements QuestionIT 1. What is the law of conservation of mass? Mass of reactants = mass products. 2. Why might some reactions appear to show a change in mass? A reactant or a product is a gas. 3. Give two examples of a reaction where a change in mass may appear to take place.

Metal reacting with oxygen or an acid. Thermal decomposition. 4. Balance the following equations: a) 2H2 + O2 2H2O b) Ca + 2HCl CaCl2 + H2 c) 2Li + 2H2O 2LiOH + H2 d) 4NH3 + 5O2 4NO + 6H2O e) 4K + O2 2K2O Chemical measurements QuestionIT 1. How many atoms and elements are in the compound sodium aluminate, NaAl(OH)4? Four elements and ten atoms. 2. What do the following formulae tell you? a) 2HCl Two molecules of hydrogen chloride. Each molecule contains one hydrogen atom and one chlorine atom

b) Cl2 One molecule of chlorine made of two atoms. 3. An aqueous solution of hydrogen peroxide (H2O2) decomposes to form water and oxygen. a) Write a balanced symbol equation for this reaction. Include the state symbols. 2H2O2 (aq) 2H2O (l) + O2(g) Chemical measurements QuestionIT b) Why does the water, produced during the reaction, have a lower mass than the original hydrogen peroxide? Because the oxygen gas produced during the reaction escaped into the atmosphere. LearnIT! KnowIT!

Chemical Measurements PART 2 Relative formula mass Chemical measurements - PART 2 Relative formula mass (Mr) of a compound is the sum of the relative atomic masses of the atoms in the numbers shown in the formula. The relative atomic masses can be found in the periodic table NaCl H2O (1 x Na + 1 x Cl) Ar: Na (23) Cl (35.5)

(2 x H + 1 x O) Ar: H (1) O (16) Mr = 23 + 35.5 = 58.5 Mr = (1 x 2) + 16 = 18 H2SO4 (2 x H + 1 x S + 4 x O) Ar: H (1) S (32) O (16) Al2(SO4)3 (2 x Al + 3 x S + 12 x O) Ar: Al (27) S (32) O (16) Mr = (1x2) + 32 + (16x4) = 98

Mr = (27x2) + (32x3) + (16x12) = 342 In a balanced chemical equation, the sum of the relative formula masses of the reactants equals the sum of the relative formula masses of the products. For example: 2Mg + O2 2MgO (2x24) + (2x16) 2 x (24+16) 80 80 QuestionIT! Chemical Measurements PART 2 Relative formula mass Chemical measurements QuestionIT 1. What is the relative formula mass of a compound? 2. What is the relative formula mass of:

a) MgCl2 b) C6H12O6 3. What can be said about the sum of the relative formula masses of the reactants and products of a reaction? 4. Why can you have relative atomic masses which are not whole numbers e.g. chlorine is 35.5? AnswerIT! Chemical Measurements PART 2 Relative formula mass Chemical measurements QuestionIT 1. What is the relative formula mass of a compound? Sum of the relative atomic masses of the atoms in the numbers

shown in the formula. 2. What is the relative formula mass of: a) MgCl2 95 b) C6H12O6 180 3. What can be said about the sum of the relative formula masses of the reactants and products of a reaction? In a balanced chemical equation the sum of the relative formula masses of the reactants in the quantities shown = sum of the relative formula masses of the products in the quantities shown. Chemical measurements QuestionIT 4. Why can you have relative atomic masses which are not whole numbers e.g. chlorine is 35.5? Isotopes.

LearnIT! KnowIT! Use of amount of substance PART 1 Moles (HT) Amounts of substances in equations (HT) Quantities in equations (HT) Using moles to balance equations (HT) Limiting reactants (HT)

Use of amount of substance - PART 1 Chemical amounts are measured in moles. The symbol for the unit mole is mol. The mass of one mole of a substance in grams is numerically equal to its relative formula mass. One mole of a substance contains the same number of the stated particles, atoms, molecules or ions as one mole of any other substance. The number of atoms, molecules or ions in a mole of a given substance is the Avogadro constant. The value of the Avogadro constant is 6.02 x 1023 per mole. Number of moles = mass (g) or mass (g) Ar Mr How many moles of sulfuric acid molecules are there in 4.7g of sulfuric acid (H2SO4)? Give your answer to 1 significant figure. 4.7 = 0.05 mol 98

Mass (g) = number of moles x Ar or number of moles x Mr What is the mass of 7.2 x 10-3 moles of aluminium sulfate (Al2(SO4)3)? Give your answer to 1 decimal place. 7.2 x 10-3 x 342 = 2.5g Use of amount of substance - PART 1 HIGHER TIER The masses of reactants and products can be calculated from balanced symbol equations. Chemical equations can be interpreted in terms of moles. Example: H2 + Cl2 2HCl This equation shows that one mole of hydrogen reacts with one mole of chlorine to form two moles of hydrochloric acid.

The balanced equation is useful because it can be used to calculate what mass of hydrogen and chlorine react together and how much hydrogen chloride is made. Ar: H (1) so mass of 1 mole of H2 = 2 x 1 = 2g Ar: Cl (35.5) so mass of 1 mole of Cl2 = 35.5 x2 = 71g Mr : HCl (1 + 35.5) so mass of 1 mole of HCl = 36.5g The balanced equation tells us that one mole of hydrogen reacts with one mole of chlorine to give two moles of hydrogen chloride molecules, so turning this to masses: 1 mole of hydrogen =1x2 = 2g 1 mole of chlorine = 1 x 71 = 71g

2 moles of hydrochloric acid = 2 x 36.5 = 73g Use of amount of substance - PART 1 HIGHER TIER Sodium hydroxide reacts with chlorine to make bleach: 2NaOH + Cl2 NaOCl + NaCl + H2O If you have a solution containing 100.0g of sodium hydroxide, what mass of chlorine gas do you need to convert it to bleach? Mr : NaOH (23 + 16 + 1) so mass of 1 mole of NaOH = 40g Mr : Cl2 (35.5 x 2) so mass of 1 mole of Cl2 = 71g So 100.0g of sodium hydroxide is 100/40 = 2.5 moles The balanced symbol equation tells us that for every two moles of sodium hydroxide, you need one mole of chlorine to react with it. So you need 2.5/2 = 1.25 moles of chlorine One mole of chlorine is 71g, so you will need 1.25 x 71g = 88.75g of chlorine to react

with 100.0g of sodium hydroxide. Use of amount of substance - PART 1 HIGHER TIER The balancing numbers in a symbol equation can be calculated from the masses of reactants and products by converting the masses in grams to amounts in moles and converting the number of moles to simple whole number ratios. 8.5g of sodium nitrate (NaNO3) is heated until its mass is constant. 6.9g of sodium nitrite (NaNO2) and 1.6g of oxygen gas (O2) is produced. NaNO3 NaNO2 + O2 Mr: NaNO3 = 23 + 14 + (16x3) = 85 Number of moles = mass (g) Mr: NaNO2 = 23 + 14 + (16x2) = 69 Mr: O2 = 16x2 = 32 M

Then to convert masses to moles use: NaNO3 : NaNO2 : r O2 Moles of NaNO3 = 8.5/85 = 0.1 mol 0.01 : 0.01 : 0.05 Moles of NaNO2 = 6.9/69 = 0.1 mol Moles of O2 = 1.6/32 = 0.05 mol Dividing the ratio by the smallest number gives 2:2:1 2NaNO3 2NaNO2 + O2 Use of amount of substance - PART 1 HIGHER TIER In a chemical reaction involving two reactants, it is common to use an excess of one of the reactants to ensure that all the reactant is used up. The reactant that is completely used up is called the limiting reactant because it limits the amount of products.

4.8g of magnesium ribbon reacts with 7.3g of HCl. Which is the limiting reactant? Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) Ar: Mg (24) and Ar: Cl (35.5) 4.8g of Mg = 4.8/24 moles = 0.2 mol 7.3g of HCl = 7.3/36.5 moles = 0.2 mol From the balanced equation: 1 mole of Mg reacts with 2 moles of HCl, therefore 0.2 mol of Mg will need 0.4 mol of HCl to react completely, there is only 0.2 mol of HCl, so the HCl is the limiting reactant. QuestionIT! Use of amount of substance PART 1

Moles (HT) Amounts of substances in equations (HT) Quantities in equations (HT) Using moles to balance equations (HT) Limiting reactants (HT) Use of amount of substance (HT) QuestionIT 1. What is meant by the term mole? 2. What is the symbol for the unit mole? 3. What does Avogadros constant tell us? 4. What is the value for Avogadros constant? 5. How many atoms in 1 mole of carbon? Use of amount of substance (HT) QuestionIT

6. How many atoms in 1 mole of chlorine gas, Cl2? 7. What can the following equation tell us about the number of moles of each substance? Mg + 2HCl MgCl2 + H2 8. What is meant by the term limiting reactant? Use of amount of substance (HT) QuestionIT 9. How many moles of helium atoms are there in 0.04g of helium? 10. What is the mass of 20 moles of calcium carbonate CaCO3? Answer in kg. 11. Calcium carbonate decomposes to calcium oxide in a kiln in the following reaction: CaCO3 CaO + CO2 Calculate the mass of calcium oxide that can be produced when

300 tonnes of calcium carbonate is heated. Use of amount of substance (HT) QuestionIT 12. 0.10g of hydrogen reacts with 3.55g of chlorine to produce 3.65g of hydrogen chloride. Use this information to work out the balancing numbers for hydrogen chloride. H2 + Cl2 ___HCl Use of amount of substance (HT) QuestionIT 13. If 4.95 g of ethene (C2H4) are combusted with 3.25 g of oxygen, what is the limiting reactant? C2H4 + 3O2 2CO2 + 2H2O AnswerIT! Use of amount of

substance PART 1 Moles (HT) Amounts of substances in equations (HT) Quantities in equations (HT) Using moles to balance equations (HT) Limiting reactants (HT) Use of amount of substance (HT) QuestionIT 1. What is meant by the term mole?

A measure of the chemical amount of a substance. 2. What is the symbol for the unit mole? mol 3. What does Avogadros constant tell us? Number of atoms, molecules or ions in a mole of a substance. 4. What is the value for Avogadros constant? 6.02 x 1023 per mol 5. How many atoms in 1 mole of carbon? 6.02 x 1023 Use of amount of substance (HT) QuestionIT 6. How many atoms in 1 mole of chlorine gas, Cl2? 6.02 x 1023 7. What can the following equation tell us about the number of moles of each substance? Mg + 2HCl MgCl2 + H2

1 mole of magnesium reacts with 2 moles of hydrochloric acid to form 1 mole of magnesium chloride and 1 mole of hydrogen. 8. What is meant by the term limiting reactant? A reactant in a reaction which is completely used up when the other reactant is in excess. Use of amount of substance (HT) QuestionIT 9. How many moles of helium atoms are there in 0.04g of helium? 0.04/4 = 0.01mol 10. What is the mass of 20 moles of calcium carbonate CaCO3? Answer in kg. 40+12+(16x3) = 100 100 x 20 = 2,000g = 2kg. 11. Calcium carbonate decomposes to calcium oxide in a kiln in the following reaction: CaCO3 CaO + CO2 Calculate the mass of calcium oxide that can be produced when

300 tonnes of calcium carbonate is heated. Relative formula mass of calcium carbonate = 100 = 100g Relative formula mass of calcium oxide = 56 = 56g 100 tonnes of calcium carbonate makes 56 tonnes of calcium oxide so 300 tonnes make 168 tonnes Use of amount of substance (HT) QuestionIT 12. 0.10g of hydrogen reacts with 3.55g of chlorine to produce 3.65g of hydrogen chloride. Use this information to work out the balancing numbers for hydrogen chloride. H2 + Cl2 ___HCl Mr : H 2 = 1 x 2 = 2 Mr: Cl2 = 35.5 x 2 = 71 Mr: HCl = 1 + 35.5 = 36.5 Then to convert masses to moles use: Moles of H2 = 0.10/2 = 0.05 mol

Moles of Cl2 = 3.55/71 = 0.05 mol Moles of HCl = 3.65/36.5 = 0.1 mol Dividing the ratio by the smallest number gives 1:1:2 H2 + Cl2 2HCl Use of amount of substance (HT) QuestionIT 13. If 4.95 g of ethene (C2H4) are combusted with 3.25 g of oxygen, what is the limiting reactant? C2H4 + 3O2 2CO2 +2H2O Mr: C2H4 = 28 Mr: O2 = 32 4.95/28 = 0.177 mol 3.25/32 = 0.102 mol From the equation: 1 mole of ethene reacts with 3 moles of oxygen. In this case 0.177 mol of ethene will need 0.53 mol of oxygen to react, which we do not have, so oxygen is the limiting

factor. LearnIT! KnowIT! Use of amount of substance PART 2 Concentration of solutions Calculations - PART 2 Chemists quote the amount of substance (solute) dissolved in a certain volume of the solution. The units used to express the concentration can be grams per decimetre cubed (g/dm3). A decimetre (1dm 3) cubed is equal to 1000cm3. The blackcurrant juice is getting more concentrated the darker colour indicates more squash is in the

same volume of its solution If you know the mass of the solute dissolved in a certain volume of solution, you can work out the concentration using: Concentration = (g/dm ) 3 amount of solute (g) Volume of solution (dm ) 3 Remember if you are using cm3 to multiply the volume by 1000 to covert to dm3 Example 1:

50g of sodium hydroxide is dissolved in water to make up 200cm3 . What is the concentration in dm3? 50g/200cm3= 0.25g/cm3 0.25g/cm3 x 1000 = 250g/dm3 Calculations - PART 2 Example 2: A solution of sodium chloride has a concentration of 200g/dm3. What is the mass of sodium chloride in 700cm 3 of solution? Convert 700cm3 into dm3 700/1000 = 0.7 dm3 Then rearrange the equation amount of solute = concentration x volume of solution

(g) (g/dm3) (dm3) 200g/dm3 x 0.7 dm3 = 140g HIGHER: You can increase the concentration of an aqueous solution by: Adding more solute and dissolving it in the same volume of its solution. Evaporating off some of the water from the solution so you have the same mass of solute in a

smaller volume of solution. QuestionIT! Use of amount of substance PART 2 Concentration of solutions Use of amount of substance QuestionIT 1. What units can be used for the concentration of a solution? 2. What does dm3 mean? 3. Give the equation for calculating concentration from the mass of substance and volume of solution. 4. HT Only: How can you increase the concentration of an aqueous solution?

Use of amount of substance QuestionIT 5. Calculate the concentration in g/dm3 for 50g of sodium chloride in 2.5 dm3 of water. 6. Calculate the concentration in g/dm3 of 1.4g of potassium carbonate in 855cm3 of water. 7. A teacher has a solution of lithium fluoride with a concentration of 72.6g/dm3. Calculate the mass of lithium fluoride dissolved in 25.0cm3 of solution. AnswerIT! Use of amount of substance PART 2

Concentration of solutions Use of amount of substance QuestionIT 1. What units can be used for the concentration of a solution? g/dm3 2. What does dm3 mean? 1000cm3 3. Give the equation for calculating concentration from the mass of substance and volume of solution. Concentration = mass volume 4. HT Only: How can you increase the concentration of an aqueous solution? Add more solute and dissolve in the same volume of water; evaporate off some of the water/decrease the volume of water Use of amount of substance QuestionIT

5. Calculate the concentration in g/dm3 for 50g of sodium chloride in 2.5dm3 of water. 50/2.5 = 20g/dm3 6. Calculate the concentration in g/dm3 of 1.4g of potassium carbonate in 855cm3 of water. (1.4/855) x 1000 = 1.64 g/dm3 7. A teacher has a solution of lithium fluoride with a concentration of 72.6g/dm3. Calculate the mass of lithium fluoride dissolved in 25.0cm3 of solution. 25cm3 = 0.025dm3 72.6 x 0.025 = 1.8g LearnIT! KnowIT! Yield and atom economy

CHEMISTRY ONLY Percentage yield Atomy economy Yield and atom economy - CHEMISTRY ONLY Even though no atoms are gained or lost in a chemical reaction, it is not always possible to obtain the calculated amount of product because: The reaction may not go to completion because it is reversible Some of the product may be lost when it is separated from the reaction mixture Some of the reactants may react in ways different to the expected reactions The amount of product obtained is known

as the yield. The theoretical yield is the maximum calculated amount of a product that could be formed from a given amount of reactants. The actual yield is the actual amount of product obtained from a chemical reaction. Yield and atom economy - CHEMISTRY ONLY When compared with the maximum theoretical amount as a percentage, it is called the percentage yield and is calculated as: Percentage yield = mass of product actually made x 100 maximum theoretical mass of product A piece of sodium metal is heated in

chlorine gas. A maximum theoretical mass of 10g for sodium chloride was calculated, but the actual yield was only 8g. Calculate the percentage yield. HIGHER: 200g of calcium carbonate is heated. It decomposes to make calcium oxide and carbon dioxide. Calculate the theoretical mass of calcium oxide made. CaCO3 CaO + CO2 Percentage yield = 8/10 x 100 = 80% This means the percentage yield is 80%

Mr of CaCO3 = 40 + 12 + (16x3) = 100 Mr of CaO = 40 + 16 = 56 100g of CaCO3 would make 56 g of CaO So 200g would make 112g Yield and atom economy - CHEMISTRY ONLY The atom economy (atom utilisation) is a measure of the amount of starting materials that end up as useful products. It is important for sustainable development and for economic reasons to use reactions with high atom economy. The percentage atom economy is calculated using a balanced equation for the reaction as follows: Relative formula mass of desired product from equation x 100 Example:

Sum of relative formula mass of all reactants from equation Calculate the atom economy for making hydrogen by reacting zinc with hydrochloric acid: Zn + 2HCl ZnCl2 + H2 Mr of H2 = 1 + 1 = 2 Mr of ZnCl2 = 65 + 35.5 + 35.5 = 136 Atom economy = 2136 + 2 100 = 2138 100 = 1.45% This method is unlikely to be chosen as it has a low atom economy. The less waste there is, the higher the atom economy, the less materials are wasted, less energy used, so making the process more economic, 'greener' and sustainable. QuestionIT! Yield and atom economy PART 1

CHEMISTRY ONLY Percentage yield Atomy economy Yield and atom economy (Chemistry only) QuestionIT 1. What is meant by the term yield? 2. What is the equation for calculating percentage yield? 3. Give 2 reasons why it is not always possible to obtain the expected amount of product from a reaction. 4. What is meant by the term atom economy? 5. Why is it important to use reactions with high atom economy? 6. What is the equation for calculating the percentage atom economy from a balanced chemical equation? Yield and atom economy (Chemistry only) QuestionIT

7. Magnesium is heated in air to make magnesium oxide. Suggest why the actual yield might be less than the maximum theoretical yield. 8. In the neutralisation of sulfuric acid with sodium hydroxide, the theoretical yield from 13.8g of sulfuric acid is 20g. In a synthesis, the actual yield is 17.4g. What is the percentage yield for this synthesis? Yield and atom economy (Chemistry only) QuestionIT 9. Calculate the atom economy for making hydrogen from methane and steam. CH4 + 2H2O CO2 + 4H2 AnswerIT! Yield and

atom economy PART 1 CHEMISTRY ONLY Percentage yield Atomy economy Yield and atom economy (Chemistry only) QuestionIT 1. What is meant by the term yield? Amount of product obtained. 2. What is the equation for calculating percentage yield? % yield = x 100 3. Give 2 reasons why it is not always possible to obtain the expected amount of product from a reaction. Reaction may not go to completion as it is reversible; some product may be lost; some reactants may react differently to expected.

4. What is meant by the term atom economy? Measure of the amount of starting materials that end up as useful products. Yield and atom economy (Chemistry only) QuestionIT 5. Why is it important to use reactions with high atom economy? Sustainable development; less waste products produced; economically viable; cheaper. 6. What is the equation for calculating the percentage atom economy from a balanced chemical equation? Atom economy = x 100 Yield and atom economy (Chemistry only) QuestionIT 7. Magnesium is heated in air to make magnesium oxide. Suggest why the actual yield might be less than the maximum theoretical

mass. Magnesium nitride is formed as well as the magnesium oxide expected/some of the oxide might escape as smoke/not all the magnesium reacts. 8. In the neutralisation of sulfuric acid with sodium hydroxide, the theoretical mass from 13.8g of sodium sulfate is 20g. In a synthesis, the actual yield is 17.4g. What is the percentage yield for this synthesis? Percentage yield = (actual yield theoretical mass) 100 Percentage yield = (17.4 20) x 100 = 87% Yield and atom economy (Chemistry only) QuestionIT 9. Calculate the atom economy for making hydrogen from methane and steam. CH4 + 2H2O CO2 + 4H2 Mr of H2O = (1 x 2) + 16 = 18

18 x 2 = 36 Mr of CH4 = 12 + (1 x 4) = 16 Atom economy = 4 236 + 16 100 = 852 100 = 15.4% LearnIT! KnowIT! Quantities CHEMISTRY ONLY Moles of solution and gases (HT) Quantities CHEMISTRY ONLY Higher The concentration of a solution is the amount of solute per volume of solution. Chemists measure concentration in moles per cubic decimetre (mol/dm3).

Concentration = (mol/dm3) amount (mol) volume (dm3) Example 1: Example 2: What is the concentration of a solution that has 35.0g of solute in 0.5dm3of solution? Calculate the mass of magnesium chloride (MgCl2) if there is 1 dm3 of a 1mol/ dm3 solution.

35/0.5 = 70 g/dm3 Mass of 1 mole of magnesium chloride = 24 + (35.5 2) = 95 g So there are 95 g of magnesium chloride in 1 dm3 of a 1 mol/dm3 solution. Quantities - CHEMISTRY ONLY If the volumes of two solutions that react completely are known and the Higher concentrations of one solution is known, the concentration of the other solution can be calculated. 2NaOH(aq) + H2SO4(aq) Na2S04(aq) + 2H2O(l) It takes 12.20cm3 of sulfuric acid to neutralise 24.00cm3 of sodium hydroxide solution, which has a concentration of 0.50mol/dm 3. Calculate the concentration of the sulfuric acid in g/dm 3 0.5 mol/dm3 x (24/1000) dm3 = 0.012 mol of NaOH

The equation shows that 2 mol of NaOH reacts with 1 mol of H 2SO4, so the number of moles in 12.20cm3 of sulfuric acid is (0.012/2) = 0.006 mol of sulfuric acid Calculate the concentration of sulfuric acid in mol/ dm 3 0.006 mol x (1000/12.2) dm3 =0.49mol/dm3 Calculate the concentration of sulfuric acid in g/ dm 3 H2SO4 = (2x1) + 32 + (4x16) = 98g 0.49 x 98g = 48.2g/dm3 Quantities - CHEMISTRY ONLY Higher Equal amounts of moles or gases occupy the same volume under the same conditions of temperature and pressure. The volume of one mole of any gas at room temperature and pressure (rtp) (20C and 1 atmospheric pressure) is 24 dm3. You can calculate the volume of a gas at room temperature and pressure from its mass and relative formula mass using the equation: Number of moles = mass

relative formula mass Volume of gas at rtp = moles x 24 You can calculate the volumes of gaseous reactants and products from a balanced equation and a given volume of a gaseous reactant or product using the following equation: Volume of gas at rtp = number of moles x molar mass volume (24 dm3) What is the volume of 3.5g of hydrogen? Ar : H (1) Mr : H2 = 2 1 mole in g = 2g 3.5/2 = 1.75 mol

Quantities - CHEMISTRY ONLY Higher 6g of a hydrocarbon gas had a volume of 4.8 dm3. Calculate its molecular mass. 1 mole = 24 dm3, so 4.8/24 = 0.2 mol Mr = 6 / 0.2 = 30 if 6g = 0.2 mol, 1 mol equals 30 g volume H2 = 1.75 x 24 = 42 dm3 What mass of magnesium carbonate is needed to make 6 dm3 of carbon dioxide? What is the volume of 11.6 MgCO3(s) + H2SO4(aq) MgSO4(aq) + H2O(l) +CO2(g) g of 1 mole = 24 dm3, 6 dm3 is equal to 6/24 = 0.25 mol butane (C4H10) gas at RTP? of gas From the equation, 1 mole of MgCO3 produces 1

Mr : (4 x 12) + (10 x 1) = 58 mole of CO2, which occupies a volume of 24 dm3. 11.6/58 = 0.20 mol so 0.25 moles of MgCO3 is needed to make 0.25 volume = 0.20 x 24 = 4.8 mol of CO2 3 dm Mr : MgCO3 = 24 + 12 + (3 x 16) = 84, Mass of MgCO3 = 0.25 x 84 = 21g QuestionIT! Quantities CHEMISTRY ONLY Moles of solution and gases (HT)

Quantities (Chemistry only) QuestionIT 1. What are the units for concentration? 2. What is the equation for the calculation of concentration from the moles and volume of solution? 3. What can be said about equal amounts of moles of gases and the volume they occupy? 4. What is meant by RTP? 5. What are the values for RTP? Quantities (Chemistry only) QuestionIT 6. What is the concentration, in g/ dm3, of a solution that has 40g of solute in 2dm3 of solution? 7. Calculate the concentration in mol/dm3 of a solution that has 0.75 mol of an acid in 3dm3 of solution.

Quantities (Chemistry only) QuestionIT 8. It takes 28.0cm3 of potassium hydroxide to neutralise 25.00cm3 of nitric acid at a concentration of 0.50 mol/dm3. HNO3 + KOH KNO3 + H2O 9. Calculate the concentration of the potassium hydroxide. What is the volume of 4.5g of oxygen? 10. Calculate the number of moles of hydrogen that occupy 6dm3 at RTP. AnswerIT! Quantities CHEMISTRY ONLY Moles of solution and gases

(HT) Quantities (Chemistry only) QuestionIT 1. What are the units for concentration? mol/dm3 (g/dm3) 2. What is the equation for the calculation of concentration from the moles and volume of solution? Concentration = 3. What can be said about equal amounts of moles of gases and the volume they occupy? Equal amounts of moles of gases occupy the same volume under the same conditions of temperature and pressure. Quantities (Chemistry only) QuestionIT 4. What is meant by RTP?

Room temperature and pressure 5. What are the values for RTP? 20oC ; 1 atmosphere pressure Quantities (Chemistry only) QuestionIT 6. What is the concentration, in g/ dm3, of a solution that has 40g of solute in 2dm3 of solution? Concentration = mass volume = 40 g 2 dm3 = 20 g/dm3 7. Calculate the concentration in mol/dm3 of a solution that has 0.75 mol of an acid in 3dm3 of solution 0.75 mol/3 dm3 = 0.25 mol/dm3 Quantities (Chemistry only) QuestionIT 8. It takes 28.0cm3 of potassium hydroxide to neutralise 25.00cm3 of nitric acid at a concentration of 0.50 mol/dm3.

HNO3 + KOH KNO3 + H2O Calculate the concentration of the potassium hydroxide. Number of moles of nitric acid = concentration volume = 0.5 mol/dm3 (25 1000) dm3 = 0.0125 mol The equation for the reaction shows that 1 mole of potassium hydroxide reacts with 1 mole of nitric acid. So there is 0.0125 mol of KOH in 28 cm3 of solution. So the concentration of KOH in mol/dm3 = number of moles volume = 0.0125 mol (28 1000) dm3 = 0.45 mol/dm3 Quantities (Chemistry only) QuestionIT 9. What is the volume of 4.5g of oxygen? Ar : O (16) Mr : O2 = 32 1 mole in g = 32g 4.5/32 = 0.14 mol

Volume O2 = 0.14 x 24 = 3.38 dm3 10. Calculate the number of moles of hydrogen that occupy 6dm3 at RTP. Number of moles = 6 24 = 0.25 mol

Recently Viewed Presentations

  • Modular Origami Cube - Murray State University

    Modular Origami Cube - Murray State University

    Modular Origami Cube Making a Cube You will need 6 squares of paper. You will need at least 3 different colors of squares (e.g., 2 red squares, 2 yellow squares, and 2 blue squares). Questions for discussion: What are the...
  • Financial Savings to the Town of (Insert Name)

    Financial Savings to the Town of (Insert Name)

    The National Volunteer Fire Council (NVFC) is the leading nonprofit membership association representing the interests of the volunteer fire, EMS, and rescue services. Organized in 1976, the NVFC serves as the voice of the volunteer fire and emergency services in...
  • Corporate Powerpoint Template - Lincoln Repository

    Corporate Powerpoint Template - Lincoln Repository

    Ian Mathews and Diane Simpson. Definitional issues ... Practice Educators found the guide useful and supported adaptation of certain aspects of their practice. The guide was also used as a tool to discuss dyslexia and support needed with students.
  • East Sussex County Council March 2019 East Sussex

    East Sussex County Council March 2019 East Sussex

    Liz Rugg. Education and ISEND. Assistant Director, Education and ISEND. Fiona Wright. ORBIS. March 2019. Business Services (ORBIS) ... Nick Skelton. Economy Division. Assistant Director, Economy. James Harris. Operations and Contract Management Division.
  • Recipe for Revolution - History with Ms. Osborn

    Recipe for Revolution - History with Ms. Osborn

    Recipe for Revolution. Revolutions are major turning points in history and regardless of where they occur, some common factors are present. These causes include a great divide between the social classes, a crisis which negatively impacts the masses (famine, drought),...
  • Supreme Court Update Lisa Soronen State and Local

    Supreme Court Update Lisa Soronen State and Local

    The Court rejected the Fire District's policy argument that "applying the ADEA to small public entities risks curtailment of vital public services such as fire protection" Experience suggests otherwise. For 30 years, the Equal Employment Opportunity Commission has consistently interpreted...
  • Complexity as Fitness for Evolved Cellular Automata Update Rules

    Complexity as Fitness for Evolved Cellular Automata Update Rules

    Times New Roman Blank Presentation Microsoft Word Document Complexity as Fitness for Evolved Cellular Automata Update Rules Rewarding Complex Behavior During Computation Speeds Evolution The Computation The Density Classification Task Time-Space Diagram of CA Complex Behavior of CA Capturing Complex...
  • Marine Life - SeaSciSurf

    Marine Life - SeaSciSurf

    Marine habitats, or ocean zones, are classified by several criteria: light, distance from land, depth, bottom. Marine life is dependent on complex dynamic exchanges with its physical environment. A number of physical factors have a controlling effect on marine life....