Calculating Mass of Substances (AQA GCSE Combined Science: Trilogy: Chemistry): Flashcards
Exam code: 8464
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- FrontThe Mole
Define the mole and state the value of the Avogadro constant.
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Define the mole and state the value of the Avogadro constant.
The mole (mol) is the SI unit of amount of substance. One mole of any substance contains 6.02 × 10²³ particles (atoms, molecules, ions, or formula units). This number is the Avogadro constant.
What is the molar mass of a substance, and how is it related to Ar or Mr?
The molar mass is the mass of one mole of a substance. For an element, it equals the Ar in grams. For a compound, it equals the Mr in grams. For example, one mole of water (Mr = 18) has a molar mass of 18 g/mol.
The formula for calculating moles is: moles = ________ ÷ ________.
Rearranged to find mass: mass = ________ × ________.
The formula for calculating moles is: moles = mass ÷ Mr.
Rearranged to find mass: mass = moles × Mr.
True or False?
To calculate the moles in 3.2 g of oxygen gas, you divide by 16 (the Ar of oxygen).
False.
Oxygen gas exists as diatomic molecules (O2).
The Mr of O2 = 2 × 16 = 32, not 16. Moles = 3.2 ÷ 32 = 0.1 mol.
Dividing by 16 gives 0.2 mol, which is a common exam error.
A student calculates the moles in 14.52 kg of ammonium nitrate (NH4NO3, Mr = 80) by entering 14.52 ÷ 80 into their calculator.
What mistake have they made?
The formula n = mass ÷ Mr requires mass in grams, not kilograms.
The student must convert first: 14.52 kg = 14 520 g.
Correct answer: 14 520 ÷ 80 = 181.5 mol.
Entering 14.52 ÷ 80 gives the wrong answer of 0.18 mol.
Calculate the mass of 0.25 mol of zinc (Zn, Ar = 65):
mass = ________ × ________ = ________ g.
Calculate the mass of 0.25 mol of zinc (Zn, Ar = 65):
mass = 0.25 × 65 = 16.25 g.
Describe the three-step method for calculating the mass of a product from a given mass of reactant.
Calculate the moles of the given substance: n = mass ÷ Mr.
Use the molar ratio from the balanced equation to find the moles of the target substance.
Calculate the mass of the target substance: mass = n × Mr.
Calculate the mass of MgO produced when 6.0 g of Mg burns completely. (Ar: Mg = 24, O = 16)
2Mg + O2 → 2MgO
Step 1: mol Mg = 6.0 ÷ 24 = 0.25 mol.
Step 2: ratio Mg:MgO = 2:2 = 1:1, so mol MgO = 0.25 mol.
Step 3: mass MgO = 0.25 × (24+16) = 0.25 × 40 = 10 g.
2Al2O3 → 4Al + 3O2
The molar ratio Al2O3:Al = ________ : ________.
If 0.5 mol Al2O3 decomposes, the moles of Al produced = ________ mol.
2Al2O3 → 4Al + 3O2
The molar ratio Al2O3:Al = 2:4 (or 1:2).
If 0.5 mol Al2O3 decomposes, the moles of Al produced = 1.0 mol.
True or False?
In a reacting mass calculation, you can work in tonnes instead of grams: the answer will still be correct as long as you are consistent throughout.
True.
Reacting masses are always in proportion to the balanced equation. You can work in grams, tonnes, kg or any other mass unit, as long as you use the same unit consistently throughout all three steps of the calculation.
Explain what the molar ratio in a balanced equation represents, and how to use it in a reacting mass calculation.
The molar ratio gives the relative amounts (moles) of each substance in the reaction. In the equation 2Mg + O2 → 2MgO, the ratio Mg:MgO = 2:2 = 1:1. To use it: multiply the moles of the given substance by the ratio to get the moles of the target substance.
N2 + 3H2 → 2NH3
Calculate the mass of NH3 from 3.0 g of H2. (Mr: H2 = 2, NH3 = 17).
Step 1: mol H2 = 3.0 ÷ 2 = 1.5 mol.
Step 2: ratio H2:NH3 = 3:2, so mol NH3 = 1.0 mol.
Step 3: mass NH3 = 1.0 × 17 = 17 g.
Describe the three-step method for deducing a balanced equation from the masses of substances in a reaction.
Divide each mass by the molar mass (Mr) to find the number of moles.
Find the simplest whole number ratio (divide all values by the smallest; multiply up if any result is not a whole number).
Use these ratios as the coefficients in the balanced equation.
64 g of CH3OH reacts with 96 g of O2 to form 88 g of CO2 and 72 g of H2O.
(Mr: CH3OH = 32, O2 = 32, CO2 = 44, H2O = 18)
Deduce the balanced equation.
Moles:
CH3OH = 64÷32 = 2
O2 = 96÷32 = 3
CO2 = 88÷44 = 2
H2O = 72÷18 = 4
Ratio 2:3:2:4 gives the balanced equation:
2CH3OH + 3O2 → 2CO2 + 4H2O.
When the mole ratios from a balanced-equation calculation are not whole numbers, you must multiply ________ the values by the ________ number.
For example, ratios 1 : 2 : 2.5 become __: __ : __ after multiplying by __.
When the mole ratios from a balanced-equation calculation are not whole numbers, you must multiply all the values by the same number.
For example, ratios 1 : 2 : 2.5 become 2: 4 : 5 after multiplying by 2.
True or False?
If the mole values you calculate do not give a simple whole number ratio, you can change the formula of one of the compounds in the equation to make the numbers work.
False.
You must never alter the formula of a compound. Instead, multiply all mole values by the same number until the smallest whole number ratio is found, then use these as the coefficients in the balanced equation.
10 g of hydrogen reacts with 80 g of oxygen to form one product.
Deduce the empirical formula of the product. (Ar: H = 1, O = 16)
Moles H = 10 ÷ 1 = 10 mol
Moles O = 80 ÷ 16 = 5 mol.
Divide by smallest (5): H = 2, O = 1.
Empirical formula = H2O.
In an experiment, 10 mol of H2 react with 5 mol of O2. Use the mole ratio to write the balanced equation for this reaction.
The ratio H2:O2 is 2:1, giving the balanced equation:
2H2 + O2 → 2H2O
Define limiting reactant and excess reactant.
The limiting reactant is the substance used up first; it determines the maximum amount of product that can form.
The excess reactant is the substance remaining when the reaction stops because the limiting reactant has been fully consumed.
If you double the mass of the limiting reactant in a reaction while keeping the other reactant in excess, what happens to the mass of product formed?
The mass of product doubles. The amount of product is directly proportional to the amount of limiting reactant, so doubling the limiting reactant doubles the yield, provided the other reactant remains in excess.
To identify the limiting reactant:
Step 1: convert each reactant mass to ________ by dividing by the ________.
Step 2: write the balanced equation and find the ________ ratio.
Step 3: compare: the reactant that ________ first is the limiting reactant.
To identify the limiting reactant:
Step 1: convert each reactant mass to moles by dividing by the molar mass (Mr)
Step 2: write the balanced equation and find the molar ratio
Step 3: compare: the reactant that runs out first is the limiting reactant
True or False?
In a two-reactant system, the reactant with fewer moles in the molar ratio of the balanced equation is always the limiting reactant.
False.
The limiting reactant depends on both the mass used and the molar ratio. You must calculate the moles of each reactant from the given masses, then compare against the ratio. The reactant with a smaller coefficient is not automatically limiting.
9.2 g of Na reacts with 8.0 g of S to form Na2S. (Ar: Na = 23, S = 32)
Which is the limiting reactant?
Moles Na = 9.2÷23 = 0.40 mol
Moles S = 8.0÷32 = 0.25 mol.
Balanced equation: 2Na + S → Na2S (ratio 2:1).
To react 0.40 mol Na needs only 0.20 mol S, but 0.25 mol S is available.
Therefore S is in excess and Na is the limiting reactant.
Excess copper carbonate is added to sulfuric acid to make copper sulfate. The purpose of the excess is to ensure all the ________ reacts. The acid is the ________ reactant.
Excess copper carbonate is added to sulfuric acid to make copper sulfate. The purpose of the excess is to ensure all the acid reacts. The acid is the limiting reactant.
Define concentration of a solution and state its units.
Concentration is the amount of solute dissolved per unit volume of solution. It is measured in grams per cubic decimetre (g/dm3), where a higher concentration means more solute is dissolved in the same volume.
How do you convert a volume from cm3 to dm3, and how can you check if you have divided correctly?
Divide by 1000 (since 1 dm3 = 1000 cm3).
You can check that you have divided correctly because you always get fewer dm3 than cm3, so the number must get smaller. If your result is larger than the original cm3 value, you have used the wrong operation.
Write the concentration formula and both rearrangements (to find mass and to find volume).
Concentration (g/dm³) = mass (g) ÷ volume (dm³)
Mass (g) = concentration × volume
Volume (dm³) = mass ÷ concentration
True or False?
250 cm3 is equal to 2.5 dm3.
False.
250 cm3 = 250 ÷ 1000 = 0.25 dm3. Dividing by 1000 always makes the number smaller. A common error is dividing by 100 instead of 1000, giving the wrong answer of 2.5 dm3.
A student dissolves 0.30 g of a solute in 25 cm3 of water. Calculate the concentration in g/dm3. Show both steps.
Step 1: convert volume: 25 ÷ 1000 = 0.025 dm3.
Step 2: concentration = 0.30 ÷ 0.025 = 12 g/dm3.
A student calculates the concentration of a solution as 0.012 g/cm3.
How do they convert this to g/dm3, and why do students often lose marks at this step?
Multiply by 1000: 0.012 × 1000 = 12 g/dm3.
Students often stop after finding g/cm3 and do not carry out this final conversion, losing the last mark. Always check that your final answer is in g/dm3.
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