Topic 3 - Quantitative Chemistry Flashcards
- Relative Formula Mass
1) True or False? The Mr of a compound is always greater than the Ar of any of the elements in that compound.
2) The Ar of oxygen is 16 and the Ar of nitrogen is 14. Find the Mr of nitric oxide, NO.
3) How do you calculate the percentage mass of an element in a compound?
4) The formula of the compound sulphur hexafluoride is SF6. Calculate the Mr of sulphur hexafluoride. Ar of F = 19 and Ar of S = 32.
5) The Mr of methane, CH4, is 16. The Ar of carbon is 12. What is the percentage mass of carbon in methane?
6) Find the Mr of calcium hydroxide, Ca(OH)2. Ar of Ca = 40, Ar of O = 16, Ar of H = 1.
7) Compound X has Mr = 30 and contains 10% hydrogen by mass. Ar of H = 1. How many hydrogen atoms are there in the molecular formula of compound X?
1) True - you find the Mr by adding the Ar of each element, so it must be greater.
2) 14+16 = 30. There’s only one atom of each element in the formula, so just add the Ar values together.
3) Multiply the Ar of the element by the number of atoms of that element in the formula of the compound. Then divide by the compound’s Mr, and multiply by 100.
4) 32 + (19 x 6) = 146
5) (12 ÷ 16) x 100 = 75%
6) 40 + [2 x (16 + 1)] = 74
7) Mass of H = 10% of Mr of X = 10% of 30 = 3
So number of H atoms = 3 ÷ (Ar of H) = 3 ÷ 1 = 3
- The Mole
1) What is the definition of a mole?
2) How is the Ar or Mr of a substance related to the idea of moles?
3) Which contains more particles, a mole of water or a mole of oxygen gas?
4) What’s the formula for the number of moles in a given mass?
5) The Mr of sodium hydroxide, NaOH, is 40. How many moles are there in 500 g of NaOH?
6) 3 moles of compound A have a mass of 126 g. What is the Mr of compound A?
7) What’s the mass of 4 moles of potassium fluoride, KF? Ar of K = 39, Ar of F = 19.
8) The formula of magnesium bromide is MgBr2. Ar of Mg = 24, Ar of Br = 80. What mass of bromine is there in 0.25 moles of MgBr2.
1) One mole is 6.02 x 10^23 (the Avogadro constant) particles of a substance.
2) The value of Ar of an element or the Mr of a compound is equal to the mass in grams of 1 mole of the substance. E.g. the Mr of CO2 is 44, so 1 moles of CO2 has a mass of 44 g.
3) Neither - they both contain the same number of particles.
4) Number of moles = mass in grams ÷ Mr
5) Number of moles = mass ÷ Mr = 500 ÷ 40 = 12.5 moles
6) Mr = mass ÷ number of moles = 126 ÷ 3 = 42
7) mass = number of moles x Mr = 4 x (39+19) = 232 g.
8) 1 moles of MgBr2 contains 2 moles of Br atoms,
so mass of Br in 1 moles of MgBr2 = 2 x 80 = 160 g.
So mass of Br in 0.25 moles = 160 ÷ 4 = 40 g.
- Conservation of Mass
1) What is meant by the term ‘conservation of mass’?
2) How does the balanced symbol equation for a reaction to show that mass is conserved in that reaction?
3) A scientist places a lump of a metal in an open container on a mass balance, then forgets to put it away before locking up the lab and going on holiday for two weeks. When the scientist returns, the reading on the mass balance has gone up. Suggest an explanation for this.
4) The equation for the reaction between sulphuric acid and sodium carbonate is:
H2SO4(aq) + Na2CO3(aq) –> Na2SO4(aq) + H2O(l) + CO2(g)
This reaction was carried out in an open beaker. Predict whether the mass of the beaker and its contents will increase, decrease or stay the same during the course of the reaction. Explain your answer.
1) Conservation of mass means the total mass of the products in a reaction will always be equal to the total mass of the reactants. This is because no atoms are ever destroyed or created during a reaction.
2) In a balanced symbol equation, the sum of the relative masses of the reactants will equal the sum of the relative masses of the products. / There will be the same number of each type of atom on both sides of the equation.
3) The metal must have reacted with one or more of the gases in the air, to form a solid or liquid product. The mass of the gas that reacted was not recorded by the balance before the reaction because the system wasn’t closed.
4) The mass will decrease. Both reactants are in solution, so both their masses will be included in the mass of the beaker at the start of the experiment. However, one of the products is carbon dioxide gas, which will escape the open beaker.
- The Mole and Equations
1) The reaction between sodium and water is:
2na + 2H2O –> 2NaOH + H2
What is the ratio of the total number of moles of reactants to the total number of moles of products in this reaction.
2) Zinc reacts with oxygen to form zinc oxide, ZnO, as follows: 2Zn + O2 –> 2ZnO
Find the total number of moles of reactants needed to make 8 moles of ZnO.
3) 0.9 moles of compound A react completely with 0.6 moles of compound B to form 0.6 moles of compound C, which is the only product of this reaction. Find the balanced equation for this reaction in terms of A, B and C.
4) Compound Z decomposes to make nitrogen gas, N2, and sodium, Na. When 2 moles of Z decomposes, it produces 46 g of sodium and 84 g of nitrogen. Deduce the balanced equation for the decomposition of Z, in terms of Z, N2 and Na. Ar of Na = 23, Ar of N = 14.
1) 4:3.
2) The equation makes 2 moles of ZnO, so to make 8 moles of ZnO, multiply by 8 ÷ 2 = 4: 8Zn + 4O2 –> 8ZnO, so making 8 moles of ZnO uses 8 + 4 = 12 moles of reactants.
3) Divide all the numbers of moles by the smallest number of moles (0.6):
A: 0.9 ÷ 0.6 = 1.5 B: 0.6 ÷ 0.6 = 1 C: 0.6 ÷ 0.6 = 1
Now multiply to get them all to be whole numbers. So in this case, multiply by 2:
A: 1.5 x 2 = 3 B: 1 x 2 = 2 C: 1 x 2 = 2
So the balanced equation is: 3A + 2B –> 2C
4) 46 g of Na = 46 ÷ 23 = 2 moles
84 g of N2 = 84 ÷ (2 x 14) = 3 moles
The numbers of moles are all whole numbers, so the balanced equation is:
2Z –> 2Na + 3N2
- Limiting Reactants
1) What’s the definition of a limiting reactant?
2) What does it mean if a reactant is said to be in excess?
3) Magnesium reacts with hydrochloric acid as follows:
Mg + 2HCl –> MgCl2 + H2
How many moles of magnesium chloride, MgCl2, form when 0.4 moles of magnesium react with an excess of hydrochloric acid?
4) Methane, CH4, burns according to the following equation: CH4 + 2O2 –> CO2 + 2H2O
What mass of water is formed when 64 g of methane burns in air? Mr of CH4 = 16, Mr of H2O = 18
5) Fluorine reacts with water to form hydrogen fluoride, HF. The reaction equation is: 2F2 + 2H2O –> O2 + 4HF.
When 76 g of F2 reacts with an excess of water, what mass of hydrogen fluoride is produced? Ar of F = 19, Ar of H = 1
1) A limiting reactant is the reactant that gets completely used up in a given reaction.
2) A reactant is in excess if there is more than enough of it present to allow the limiting reactant to be fully used up.
3) 0.4 moles. The equation shows you get 1 mole of MgCl2 for each mole of Mg that reacts. The acid is in excess, so all 0.4 moles of magnesium will react to make 0.4 moles of MgCl2.
4) Number of moles in 64 g of CH4 = mass ÷ Mr = 64 ÷ 16 = 4 From the equation, each mole of CH4 makes 2 moles of H2O, so 2 x 4 = 8 moles of H2O are made.
So mass of H2O produced = moles x Mr = 8 x 18 = 144 g.
5) Mr of F2 = 2 x 19 = 38, so moles of F2 reacting = mass ÷ Mr = 76 ÷ 38 = 2. From the equation, 2 moles of F2 produces 4 moles of HF. Mr of HF = 1 + 19 = 20, so mass of HF produced = moles x Mr = 4 x 20 = 80 g.
- Volume of Gases
1) What’s the volume of one moles of any gas at room temperature and pressure?
2) What’s the formula for calculating the volume of a gas at r.t.p from its mass?
3) What volume does 1.6 kg of bromine gas, Br2, occupy at r.t.p? Ar of Br = 80.
4) The reaction between nitric oxide gas, NO, and oxygen is: 2NO + O2 –> 2NO2. What mass of nitrogen dioxide gas, NO2, is produced when 120 dm³ of NO reacts completely with oxygen at r.t.p? Ar of N = 14, Ar of O = 16
5) The equation for the decomposition of calcium carbonate is: CaCO3 –> CaO + CO2
What is the volume at r.t.p of the carbon dioxide produced when 25 g of calcium carbonate fully decomposes?
Ar of Ca = 40, Ar of C = 12, Ar of O = 16.
1) 24 dm³. The same number of moles of any two gases will always occupy the same volume as each other if they’re at the same temperature and pressure.
2) Volume of gas = mass of gas/Mr of gas x 24
3) The mass needs to be in grams. 1.6 kg x 1000 = 1600 g.
Mr of Br2 = 2 x 80 = 160.
So volume = (mass of gas ÷ Mr of gas) x 24 = (1600 ÷ 160) x 24 = 10 x 24 = 240 dm³
4) From the equation, 120 dm³ of NO will produce 120 dm³ of NO2, so mass of NO2 formed = (volume ÷ 24) x Mr of NO2 = (120 ÷ 24) x (14 + (2 x 16)) = 5 x 46 = 230 g.
5) Moles of CaCO3 = mass ÷ Mr = 25 ÷ (40 + 12 + (3 x 16)) = 25 ÷ 100 = 0.25. From the equation, 0.25 moles of CaCO3 produces 0.25 moles CO2.
Volume at r.t.p of 0.25 moles of CO2 = 0.25 x 24 = 6 dm³.
- Concentrations of Solutions
1) How do you find the concentration of a solution in g/dm³?
2) A large bottle of a popular brand of soft drink contains 130 g of sugar. The bottle holds 2 dm³ of the drink. The drink can’t be named, for legal reasons. What is the sugar concentration of the drink in g/dm³?
3) What mass of sodium sulphate is there in 100 cm³ of sodium sulphate solution if the solution has a concentration of 5 g/dm³?
4) Misty-Marie dissolved some sodium chloride in water to form a solution. She then made a second solution, using the same mass of sodium chloride, but only half as much water. How will the concentration of the second solution compare with the first?
5) Find the concentration, in mol/dm³, of the solution formed when 0.2 moles of potassium nitrate are dissolved in 250 cm³ of water.
1) Divide the mass of solute in grams by the volume of the solution in dm³.
2) Concentration = mass ÷ volume = 130 ÷ 2 = 65 g/dm³.
3) 100 cm³ = 100 ÷ 1000 = 0.1 dm³
Mass = concentration x volume = 5 x 0.1 = 0.5 g.
4) Halving the volume of the solution while keeping the mass of sodium chloride the same will double the concentration. If that’s not clear, try it with some numbers - for example, 50 g dissolved in 2 dm³ is a concentration of 25 g/dm³, but if you halve the volume and keep the mass the same, the concentration doubles to 50 g/dm³.
5) First convert the volume to dm³: 250 ÷ 1000 = 0.25 dm³
So concentration = moles ÷ volume = 0.2 ÷ 0.25 = 0.8 mol/dm³
- Concentration Calculations
1) What’s the formula for concentration in mol/dm³?
2) How many moles of solute are there in 4.0 dm³ of a 0.3 mol/dm³ solution?
3) A solution of potassium bromide, KBr, has a concentration of 0.2 mol/dm³. What is this in g/dm³? Mr of KBr = 119
4) How many grams of beryllium chloride, BeCl2, are there in 4.0 dm³ of a solution with a concentration of 0.50 mol/dm³?
Ar of Be = 9, Ar of Cl = 35.5
1) Concentration = number of moles ÷ volume of dm³
2) Number of moles = concentration x volume = 0.3 x 4.0 = 1.2 moles.
3) Mass = moles x Mr = 0.2 x 119 = 23.8 g, so concentration = 23.8 g/dm³
4) Moles of BeCl2 = concentration x volume = 0.50 x 4.0 = 2.0 moles
Mass of BeCl2 = no. of moles x Mr = 2.0 x [9+ (2 x 35.5)] = 2.0 x 80 = 160 g
- Atom Economy
1) What is atom economy?
2) What does it tell you if a reaction has an atom economy of 100%?
3) What’s the formula for calculating atom economy?
4) The following reaction can be used to produce calcium oxide, CaO: CaCO3 –> CaO + CO2
Find the atom economy of this reaction.
Ar of Ca = 40, Ar of C = 12, Ar of O = 16
5) Explain two advantages of using a reaction with a high atom economy in industry.
1) The atom economy of a reaction is the percentage of the mass of the reactants that ends up as part of the desired product.
2) All the atoms in the reactants become part of the desired product(s), so no waste is produced.
3) Atom economy = (relative formula mass of desired products ÷ relative formula mass of all reactants) x 100
4) Atom economy = [Mr of CaO ÷ Mr of CaCO3] x 100
= [(40 + 16) ÷ (40 + 16 + (12 x 3))] x 100 = [56 ÷ 100] x 100 = 56%
5) High atom economy reactions use less raw materials so are more sustainable than low atom economy reactions. There is less waste to be disposed of with high atom economy reactions than low atom economy reactions.
- Percentage Yield
1) What is the formula for working out the percentage yield of a reaction?
2) True or False? A reaction with only one product always has a 100% yield.
3) Explain why you can never get a 100% yield from a reversible reaction.
1) Percentage yield = (mass of product actually made ÷ maximum theoretical mass of product) x 100
2) False. The percentage yield doesn’t depend on how many products there are, but on how much of the product you actually produce.
3) In a reversible reaction, both the forward and backwards reaction happen at the same time, so some of the products always get converted back to the reactants and the reaction never goes to completion.
