Quantitative chemistry flashcards
13 cards on Quantitative chemistry: the key points, worked examples and the slips to watch. Swipe through them, or print them.
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Quantitative chemistry
Conservation of mass
What is the law of conservation of mass, and why can a mass seem to change in an open container?
Answer it in your head, then turn the card over.
Conservation of mass
Key points
- No atoms are made or destroyed in a reaction: they are only rearranged.
- So the total mass of the products equals the total mass of the reactants.
- A solid gets lighter if a gas it makes escapes, such as carbon dioxide from a heated carbonate.
- A solid gets heavier if it takes in a gas, such as oxygen from the air when magnesium burns.
2.4 g of magnesium makes 4.0 g of oxide
The extra mass is oxygen taken from the air.
Oxygen = 4.0 − 2.4 = 1.6 g
Watch out
A change in mass in an open flask does not break the law: look for a gas that escaped or was taken in.
Quantitative chemistry
Balancing equations
How do you balance a chemical equation?
Answer it in your head, then turn the card over.
Balancing equations
Key points
- A balanced equation has the same number of each type of atom on both sides.
- Count the atoms of each element on each side.
- Put numbers in front of formulas to even them up, one element at a time, then count again.
H₂ + O₂ → H₂O
2 oxygen atoms on the left, 1 on the right: make it 2H₂O.
Now 4 hydrogen atoms on the right, so make it 2H₂: 2H₂ + O₂ → 2H₂O
Check: hydrogen 2 × 2 = 4 on each side, oxygen 2 on each side.
Watch out
Never change a formula to balance it: H₂O₂ is a different chemical from H₂O. Only the numbers in front change.
Quantitative chemistry
Relative formula mass
How do you work out the relative formula mass (Mr) of a substance?
Answer it in your head, then turn the card over.
Relative formula mass
Key points
- Mr is the sum of the relative atomic masses (Ar) of all the atoms in the formula.
- Take each Ar from the periodic table, and count every atom the formula shows.
- A small number after a symbol counts only the atom just before it.
- Mr has no units.
Carbon dioxide, CO₂ (C = 12, O = 16)
One carbon and two oxygens
Mr = 12 + 2 × 16 = 44
Watch out
The 2 in CO₂ doubles the oxygen only, not the carbon.
Quantitative chemistry
Relative formula mass with brackets
How do you work out the Mr of a formula with brackets, such as Ca(OH)₂?
Answer it in your head, then turn the card over.
Relative formula mass with brackets
Key points
- A number after a bracket multiplies everything inside the bracket.
- Work out the bracket first, multiply it, then add the rest.
- Ca(OH)₂ has one calcium, two oxygens and two hydrogens.
Ca(OH)₂ (Ca = 40, O = 16, H = 1)
Bracket: 16 + 1 = 17
Mr = 40 + 2 × 17 = 74
Watch out
Doubling only the hydrogen gives 40 + 16 + 2 = 58: the 2 doubles the oxygen too.
Quantitative chemistry
HigherThe mole and the Avogadro constant
What is a mole, and how many particles does it hold?
Answer it in your head, then turn the card over.
The mole and the Avogadro constant
Key points
- A mole is a fixed number of particles: 6.02 × 10²³. This is the Avogadro constant.
- One mole of a substance has a mass in grams equal to its Mr (or Ar for an element).
- So 18 g of water and 44 g of carbon dioxide each hold one mole of molecules.
- Equal numbers of moles means equal numbers of particles, whatever the substance.
Molecules in 0.5 mol of water
0.5 × 6.02 × 10²³ = 3.01 × 10²³ molecules
Watch out
A mole is a count, like a dozen, not a mass: one mole of hydrogen gas is only 2 g.
Quantitative chemistry
HigherMoles from a mass
How do you work out the number of moles in a mass of a substance?
Answer it in your head, then turn the card over.
Moles from a mass
Key points
- Divide the mass in grams by the Mr.
- Rearrange to find a mass: mass = moles × Mr.
- Work out the Mr first, from the relative atomic masses.
Formulae
moles = mass ÷ Mr
mass = moles × Mr
22 g of carbon dioxide (Mr = 44)
Moles = 22 ÷ 44 = 0.5 mol
Watch out
Mass goes on top: 44 ÷ 22 = 2 is upside down.
Quantitative chemistry
HigherReacting masses
How do you use a balanced equation to work out the mass of a substance that reacts or forms?
Answer it in your head, then turn the card over.
Reacting masses
Key points
- The balancing numbers give the ratio of moles that react.
- Step 1: change the mass you know into moles (mass ÷ Mr).
- Step 2: use the ratio from the equation to find the moles of the other substance.
- Step 3: change those moles back into a mass (moles × Mr).
2Mg + O₂ → 2MgO: oxygen for 4.8 g of Mg
Moles of Mg = 4.8 ÷ 24 = 0.2
The ratio is 2 Mg : 1 O₂, so moles of O₂ = 0.2 ÷ 2 = 0.1
Mass of O₂ = 0.1 × 32 = 3.2 g
Watch out
Do not skip the ratio: the moles are only the same when the balancing numbers are the same.
Quantitative chemistry
HigherBalancing an equation from masses
How do you use reacting masses to work out the balancing numbers in an equation?
Answer it in your head, then turn the card over.
Balancing an equation from masses
Key points
- Change the mass of each reactant and product into moles.
- Divide every answer by the smallest one to get a whole-number ratio.
- Those numbers are the balancing numbers.
2.4 g Mg + 1.6 g O₂ → 4.0 g MgO
Mg: 2.4 ÷ 24 = 0.1. O₂: 1.6 ÷ 32 = 0.05. MgO: 4.0 ÷ 40 = 0.1
Divide by 0.05: 0.1 ÷ 0.05 = 2, so the ratio is 2 : 1 : 2
2Mg + O₂ → 2MgO
Watch out
Use the Mr of the molecule as written: oxygen gas is O₂, Mr 32, not 16.
Quantitative chemistry
HigherLimiting reactants
What is a limiting reactant, and how does it decide the amount of product?
Answer it in your head, then turn the card over.
Limiting reactants
Key points
- The limiting reactant is completely used up first, so it stops the reaction.
- The other reactants are in excess: some is left over.
- The amount of product depends only on the limiting reactant.
- Find it by working out how many moles of each the equation needs.
2Mg + O₂ → 2MgO, 0.3 mol Mg, 0.1 mol O₂
0.1 mol O₂ needs 2 × 0.1 = 0.2 mol Mg, so 0.3 − 0.2 = 0.1 mol Mg is left over.
Oxygen is limiting: 2 × 0.1 = 0.2 mol of MgO forms.
Watch out
Adding more of the reactant in excess makes no more product: only more of the limiting one does.
Quantitative chemistry
Concentration in g/dm³
How do you calculate the concentration of a solution in grams per dm³?
Answer it in your head, then turn the card over.
Concentration in g/dm³
Key points
- Concentration is the mass of solute dissolved in a given volume of solution.
- The volume must be in dm³: divide a volume in cm³ by 1000.
- 1 dm³ = 1000 cm³, which is one litre.
Formula
concentration (g/dm³) = mass of solute (g) ÷ volume (dm³)
6 g of salt in 200 cm³ of solution
Volume = 200 ÷ 1000 = 0.2 dm³
Concentration = 6 ÷ 0.2 = 30 g/dm³
Watch out
Dividing by 200 instead of 0.2 gives 0.03 g/dm³: change cm³ to dm³ first.
Quantitative chemistry
Mass, volume and concentration
How do the mass of solute and the volume change a solution's concentration, and how do you find the mass dissolved?
Answer it in your head, then turn the card over.
Mass, volume and concentration
Key points
- More solute in the same volume: more concentrated.
- The same solute in less solution, such as after evaporating some water: more concentrated.
- Adding more water dilutes it: less concentrated.
- Rearrange to find the mass: mass = concentration × volume.
Mass in 250 cm³ of a 40 g/dm³ solution
Volume = 250 ÷ 1000 = 0.25 dm³
Mass = 40 × 0.25 = 10 g
Watch out
Pouring some solution away does not change its concentration: the solute and the water both go down together.
Quantitative chemistry
Chemistry onlyPercentage yield
How do you calculate a percentage yield, and why is it almost always below 100%?
Answer it in your head, then turn the card over.
Percentage yield
Key points
- The theoretical yield is the most product the balanced equation says you can make.
- The actual yield is what you really collect.
- Yields fall short: the reaction may not finish, some product is lost when it is separated, or side reactions make other products.
Formula
percentage yield = (actual yield ÷ theoretical yield) × 100
16 g collected, 20 g possible
Percentage yield = 16 ÷ 20 × 100 = 80%
Watch out
A yield over 100% means the division is upside down: 20 ÷ 16 × 100 = 125% cannot happen.
Quantitative chemistry
Chemistry onlyAtom economy
What is atom economy, and why do chemists want it to be high?
Answer it in your head, then turn the card over.
Atom economy
Key points
- Atom economy is the percentage of the mass of the reactants that ends up in the product you want.
- A high atom economy wastes fewer atoms, which saves resources and money and makes less waste.
- A reaction with only one product has an atom economy of 100%.
Formula
atom economy = (Mr of wanted product ÷ total Mr of all reactants) × 100
CaCO₃ → CaO + CO₂, wanting CaO
Mr: CaCO₃ = 100, CaO = 56
Atom economy = 56 ÷ 100 × 100 = 56%
Watch out
Atom economy comes from the equation, not the experiment: it is not the same as percentage yield.
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