2.1.3e,g - Reacting masses, concentrations and stoichiometry

2.1.3e,g - Reacting masses, concentrations and stoichiometry

Chemists use amount of substance, measured in mol, as the bridge between particles and measurable quantities. In this lesson you will use mass, gas volume, and solution concentration to find amounts in mol, then use balanced equations to link the amounts of different substances in a reaction. This is the calculation route behind reacting masses, gas volumes and many later quantitative chemistry questions.

The mole bridge for calculations

In a reaction calculation, the central quantity is usually amount of substance, n, in mol. Mass, gas volume and solution concentration are different ways of measuring a substance, but each can be converted into mol.

The three main conversion routes are:

Mass and amount

n=mMn = \frac{m}{M}

where n is amount in mol, m is mass in g, and M is molar mass in g mol^-1.

Gas volume at RTP

n=V24.0n = \frac{V}{24.0}

where V is gas volume in dm^3 at room temperature and pressure, RTP.

Solution concentration

n=cVn = cV

where c is concentration in mol dm^-3 and V is solution volume in dm^3.

The volume unit is a common source of lost marks:

1000 cm3=1.000 dm31000\ \text{cm}^3 = 1.000\ \text{dm}^3

So:

volume in dm3=volume in cm31000\text{volume in dm}^3 = \frac{\text{volume in cm}^3}{1000}

For concentrations, OCR questions may use either mol dm^-3 or g dm^-3.

Concentration in g dm^-3

concentration in g dm−3=mass of solute in gvolume of solution in dm3\text{concentration in g dm}^{-3} = \frac{\text{mass of solute in g}}{\text{volume of solution in dm}^3}

You can connect the two concentration units using molar mass:

concentration in g dm−3=concentration in mol dm−3×M\text{concentration in g dm}^{-3} = \text{concentration in mol dm}^{-3} \times M

For example, a 0.200 mol dm^-3 solution of NaOH has mass concentration:

0.200×40.0=8.00 g dm−30.200 \times 40.0 = 8.00\ \text{g dm}^{-3}

because the molar mass of NaOH is 40.0 g mol^-1.

Balanced equations give mole ratios

A balanced equation tells you the ratio of amounts in mol. It does not directly give a ratio of masses.

For example:

2Mg(s)+O2(g)→2MgO(s)2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)

This means:

  • 2 mol of Mg react with 1 mol of O2.
  • 2 mol of Mg produce 2 mol of MgO.
  • The mole ratio Mg : MgO is 2 : 2, which simplifies to 1 : 1.

The reliable route is:

  1. Convert the given quantity to amount in mol.
  2. Use the balanced-equation mole ratio.
  3. Convert the required amount into the requested quantity.

Worked example: mass of product from mass of reactant

Calculate the mass of magnesium oxide formed when 1.20 g of magnesium reacts completely with oxygen.

Equation:

2Mg(s)+O2(g)→2MgO(s)2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)

Use M(Mg) = 24.3 g mol^-1 and M(MgO) = 24.3 + 16.0 = 40.3 g mol^-1.

First find the amount of Mg:

n(Mg)=1.2024.3=0.04938... moln(\text{Mg}) = \frac{1.20}{24.3} = 0.04938...\ \text{mol}

The ratio Mg : MgO is 2 : 2, so the amount of MgO is the same:

n(MgO)=0.04938... moln(\text{MgO}) = 0.04938...\ \text{mol}

Now convert amount of MgO to mass:

m(MgO)=nM=0.04938...×40.3=1.99 gm(\text{MgO}) = nM = 0.04938... \times 40.3 = 1.99\ \text{g}

To three significant figures, the mass of MgO is:

1.99 g1.99\ \text{g}

Use the equation ratio between amounts in mol, then convert into the final unit. Never apply the equation ratio directly to masses.

Gas volume calculations at RTP

For gases at RTP, the data sheet value is:

1 mol gas=24.0 dm31\ \text{mol gas} = 24.0\ \text{dm}^3

This route only applies when the question is using molar gas volume at RTP. The ideal gas equation pV = nRT belongs to a separate calculation method and is not needed in this lesson.

Worked example: mass to gas volume

Calcium carbonate decomposes on heating:

CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

Calculate the volume of carbon dioxide formed at RTP from 0.500 g of calcium carbonate.

Use M(CaCO3) = 40.1 + 12.0 + (3 x 16.0) = 100.1 g mol^-1.

First convert the mass of CaCO3 to amount:

n(CaCO3)=0.500100.1=0.004995... moln(\text{CaCO}_3) = \frac{0.500}{100.1} = 0.004995...\ \text{mol}

The equation ratio CaCO3 : CO2 is 1 : 1, so:

n(CO2)=0.004995... moln(\text{CO}_2) = 0.004995...\ \text{mol}

Convert amount of CO2 to gas volume at RTP:

V(CO2)=n×24.0=0.004995...×24.0=0.1199... dm3V(\text{CO}_2) = n \times 24.0 = 0.004995... \times 24.0 = 0.1199...\ \text{dm}^3

To three significant figures:

V(CO2)=0.120 dm3V(\text{CO}_2) = 0.120\ \text{dm}^3

This is also:

0.120×1000=120 cm30.120 \times 1000 = 120\ \text{cm}^3

Why the unit matters

If a gas volume is given in cm^3, convert it to dm^3 before using n = V/24.0. For example, 96.0 cm^3 = 0.0960 dm^3.

Solution volume and concentration

For solutions, concentration in mol dm^-3 tells you the amount of solute in one cubic decimetre of solution.

c=nVc = \frac{n}{V}

So:

n=cVn = cV

where volume must be in dm^3.

Worked example: reacting volumes of solutions

Silver nitrate solution reacts with sodium chloride solution:

AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq)\text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)

Calculate the volume of 0.150 mol dm^-3 sodium chloride solution needed to react exactly with 20.0 cm^3 of 0.100 mol dm^-3 silver nitrate solution.

First calculate the amount of AgNO3:

20.0 cm3=0.0200 dm320.0\ \text{cm}^3 = 0.0200\ \text{dm}^3 n(AgNO3)=cV=0.100×0.0200=0.00200 moln(\text{AgNO}_3) = cV = 0.100 \times 0.0200 = 0.00200\ \text{mol}

The equation ratio AgNO3 : NaCl is 1 : 1, so:

n(NaCl)=0.00200 moln(\text{NaCl}) = 0.00200\ \text{mol}

Rearrange n = cV to find the volume of NaCl solution:

V=ncV = \frac{n}{c} V(NaCl)=0.002000.150=0.01333... dm3V(\text{NaCl}) = \frac{0.00200}{0.150} = 0.01333...\ \text{dm}^3

Convert to cm^3:

0.01333...×1000=13.3 cm30.01333... \times 1000 = 13.3\ \text{cm}^3

Concentration in g dm^-3

If the concentration is given in g dm^-3, convert to either mass or mol depending on what the question asks.

For example, a sodium chloride solution contains 11.7 g dm^-3 of NaCl. Use M(NaCl) = 58.5 g mol^-1.

In mol dm^-3:

c=11.758.5=0.200 mol dm−3c = \frac{11.7}{58.5} = 0.200\ \text{mol dm}^{-3}

So 25.0 cm^3 of this solution contains:

n=cV=0.200×0.0250=0.00500 moln = cV = 0.200 \times 0.0250 = 0.00500\ \text{mol}

Multi-step stoichiometry

A longer question is still built from the same small moves. Read the question for:

  • the balanced equation
  • the quantity you are given
  • the substance you must find
  • the unit required in the final answer

Then follow the route:

given quantity→mol of given substance→mol of required substance→final quantity\text{given quantity} \rightarrow \text{mol of given substance} \rightarrow \text{mol of required substance} \rightarrow \text{final quantity}

Worked example: solution concentration to gas volume

Hydrogen peroxide decomposes as shown:

2H2O2(aq)→2H2O(l)+O2(g)2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)

Calculate the volume of oxygen formed at RTP when 100 cm^3 of 0.500 mol dm^-3 hydrogen peroxide decomposes completely.

Convert the solution volume:

100 cm3=0.100 dm3100\ \text{cm}^3 = 0.100\ \text{dm}^3

Calculate the amount of H2O2:

n(H2O2)=cV=0.500×0.100=0.0500 moln(\text{H}_2\text{O}_2) = cV = 0.500 \times 0.100 = 0.0500\ \text{mol}

Use the equation ratio. 2 mol of H2O2 form 1 mol of O2:

n(O2)=0.05002=0.0250 moln(\text{O}_2) = \frac{0.0500}{2} = 0.0250\ \text{mol}

Convert amount of O2 to volume at RTP:

V(O2)=0.0250×24.0=0.600 dm3V(\text{O}_2) = 0.0250 \times 24.0 = 0.600\ \text{dm}^3

So the volume of oxygen is:

0.600 dm30.600\ \text{dm}^3

or:

600 cm3600\ \text{cm}^3

Multi-step stoichiometry Summary

Stoichiometry is not a separate formula to memorise. It is the balanced-equation mole ratio placed between two unit conversions.

Explain It Back

Use this as a self-explanation check after the section above. It is for diagnosing what you can already explain, not for learning new material from scratch.