# Reacting Quantites

## Solutions, and making standard solutions

We usually measure the concentration of a solution in moles per dm3 (mol dm-3). The old term ‘Molar’ is sometimes seen on solutions – a 1 Molar solution (1M) contains 1 mole of solute per dm3 of solvent.

We can write this as: concentration (mol dm-3)  =  moles ÷ volume (in dm3)

Conversely:

moles = concentration (mol dm-3) x volume (in dm3)

Remember that 1 dm3 = 1000 cm3= 1 litre  :  Common mistake, forgetting to convert cm3 to dm3

Occasionally you may see concentrations written in units of grams per cm3 (g cm-3). Since the relationship between moles and mass is simply   mass = moles x Mr we can easily work in either set of units for concentrations:

concentration in g cm-3 = concentrtion in mol dm-3 x Mr

concentration in mol dm-3 = concentration in g dm-3 ÷ Mr

### Application

```Q: How many moles of HCl in 100cm3 of 0.2 mol dm-3 HCl?
A: moles = 0.2 × (100 ÷ 1000)    = 0.02 mol```
```Q: What is the concentration of a solution containing 0.3 mol NaOH dissolved in 500cm3 of the solution?
A: concentration = moles ÷ volume in dm3   = 0.3 ÷ 0.5  = 0.6 mol dm-3```
```Q: What is the concentration of a solution containing 5.61g of potassium hydroxide dissolved in 100cm3 of the solution?
A: First convert mass to moles of KOH       Mr = 56.1     Moles = 5.61 ÷ 56.1 = 0.100
Then use concentration = moles ÷ volume in dm3  = 0.100 ÷ 0.100   = 1.00 mol dm-3```
```Q: Calculate the mass of NaOH required to produce 250cm3 of 0.100M solution?
A: First work out moles in the solution    moles = conc x vol in dm3
= 0.100 x 0.250 = 0.0250 moles
Then work out mass of NaOH required:  Mr NaOH = 40.1
mass = moles x Mr   = 0.0250 x 40.1   = 1.00g (to 3sf)
```

### Making a Standard Solution

A standard solution is one that has a precisely known concentration. Making a standard solution involves dissolving a precisely-known mass of solute in a precisely-known volume of pure water.

The procedure to make 250 cm3 of a standard solution is:

1. Weigh out the required mass of solute using a balance with sufficient decimal places.
2. Transfer the solid to a beaker, washing any residual solid into the beaker with distilled water.
3. Add sufficient distilled water to dissolve the solid, but less than 250 cm3 , and stir with a glass rod to help the solid dissolve.
4. Once the solid has dissolved, transfer the solution using a funnel into a 250cm3  volumetric flask.
5. Wash the remaining solution from the beaker into the volumetric flask with a little distilled water, and rinse the glass rod and funnel into the volumetric flask with distilled water, ensuring that the total volume of the solution still does not exceed 250cm3 .
6. Use a pipette to bring the volume of the solution in the volumetric flask up to the graduated mark, so that the bottom of the meniscus is on the mark.
7. Stopper the volumetric flask and invert a few times to thoroughly mix the solution.

N.B.    A volumetric flask is a precision piece of glassware and must never be heated.

## Amount of substance

When carrying out reactions, we do all our calculations in moles. Moles are the correct System Internationale (SI) unit for ‘amount of substance’. We need to understand that the numbers in front of formulae in balanced equations (the stoichiometry) are numbers of moles.

### Using masses (solids, liquids or gases)

Solids are measured out by mass. Sometimes masses of pure liquids may be measured out, although volume would be more usual. Masses of gases are occasionally used.

We convert to/from moles using: moles = mass ÷ Mr       and        mass = moles × Mr

### Using volume and density (pure liquids)

For pure liquids, the density of the liquid (units: grams per cm3) is used to allow us to convert between volume and mass. We need to ensure the volume of liquid is measured in the same units as the density – cm3 is most convenient.

mass = volume (in cm3) × density,  therefore    moles = (volume × density) ÷ Mr

and   volume (cm3) = (moles x Mr) ÷ density

### Using volume and concentration (for solutions only)

As previously explained, for a solution we use: moles = concentration x volume (in dm3)

and volume (in dm3) = moles / concentration

### Using gaseous volumes

It is not very convenient to measure the mass of gases. We normally measure the amount of a gas by volume.

Avagadro’s Law:   Equal volumes of all gases, measured at the same temperature and pressure contain the same number of molecules (and hence moles).

If we know the molar gas volume at the temperature and pressure we are using, then we can use the relationships: moles = volume (in dm3)  ÷  molar gas volume (in dm3)

and   volume (in dm3) = moles  ×  molar gas volume (in dm3)

At room temperature and pressure (denoted r.t.p.) the molar volume volume (the volume of one mole of any gas) is taken to be 24.0 dm3 (which is 24,000 cm3).

Under other conditions when we don’t know the molar gas volume we make use of the ideal gas equation. This states that:

pV = nRT       where p = pressure in Pascals    [from kPa to Pa:  multiply by 103]

V = volume of gas in m3     [from cm3 to m3:  multiply by 106]

n = moles of gas

R = a constant: value is always 8.314 Jmol-1K-1

T = temperature in Kelvin   [from °C to K:  add 273]

### Application

```Q: How should 0.200 moles of magnesium be measured out?
A: Magnesium is a solid, so  mass = moles × Mr   = 0.200 × 24.3   = 4.86 g of Mg```
```Q: How should 0.200 moles of  2.00 mol dm-3 hydrochloric acid be measured out?
A: It is a solution, so  vol (in dm3) = moles ÷ conc. = 0.200 ÷ 2.00  = 0.100 dm3  (or 100cm3)```
```Q: How should 0.200 moles of ethanol (Mr = 46; density 0.79 gcm-3) be measured out?
A: For a pure liquid, vol (in cm3) = (moles × Mr) ÷ density = (0.200 x 46) ÷ 0.79  = 11.6 cm3  (to 3 s.f.)
alternatively, mass = moles × Mr   = 0.200 × 46   = 9.2 g```
```Q: How should 0.200 moles of hydrogen gas at r.t.p. be measured out?
A: vol (in dm3) = moles x molar gas volume = 0.200 x 24.0 = 4.80 dm3```
```Q: How should 0.200 moles of hydrogen gas at -50.0°C & 300kPa be measured out?
A: Convert -50.0°C  to 223K  and  300kPa to 300,000 Pa.   Rearrange pV = nRT
⇒  V (in m3) = nRT/p  = (0.200 × 8.314 × 223) ÷ 300,000   = 1.24x10-3 m3  = 1.24dm3```

## Reacting Quantity Calculations

Since we can convert various measured quantities into moles of substance, we can use balanced chemical equations to work out how many moles of other substances would react or be produced using the mole ratios of the balanced equation. We can then convert these amounts in moles to mass, volume etc. as required.

It is important to understand the idea of being in excess. When one reactant is completely used up, the reaction must stop even though there may be more of the other reactants remaining. These are said to be in excess, and it will be necessary to use the moles of the reactant which is not in excess and the balanced equation to work out how much of the reactants that are in excess actually reacted, and how much remains when the reaction is complete.

### Application

10.0g of iron is warmed gently with 150cm3 of 1.00 mol dm-3 sulphuric acid, reacting according to the equation: Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)

1. Calculate the volume of H2 produced at r.t.p.   [molar gas volume = 24.0 dm3 mol-1]
2. What would you observe during this reaction?
3. Calculate the concentration of the resulting iron(II) sulphate solution
4. Suggest how a sample of hydrated iron(II) sulphate crystals could be obtained
5. Calculate the maximum mass of iron(II) sulphate, FeSO4.7H2O that could be obtained

```1.  moles Fe (Ar = 55.8) =  mass ÷ Ar  = 10.0 ÷ 55.8 = 0.179 mol
moles H2SO4 = conc × vol in dm3  = 1.00 × 0.150  = 0.150 mol
Fe and H2SO4 react in a 1:1 mole ratio, so Fe is in excess
therefore only 0.150 mol of the Fe will react
so 0.150 mol of H2 will be produced (1:1 mole ratio Fe to H2)
vol of H2 = moles × molar volume  =  0.150 × 24.0  = 3.60 dm3```
```2.  Look at the state symbols:  Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
You would observe fizzing/bubbles/effervescence.
Some of the iron dissolves/disappears but not all.
The iron sulphate product is formed as a solution (it happens to be pale green).```
```3. Since it is a 1:1 mol ratio (Fe to FeSO4), then 0.150 mol of FeSO4 will be made
It will be dissolved in the 150cm3 volume of the solution.
Concentration = moles ÷ volume in dm3 = 0.150 ÷ 0.150 = 1.00 mol dm-3.```
```4. Filter to remove the remaining unreacted iron (1.63g of it will remain after the reaction)
Heat the solution in an evaporating dish to remove some of the water
Allow to cool and crystals to form
Filter to remove the crystals from the remaining solution
Dry the crystals (warm oven, or between filter paper)```
`5. Mr of FeSO4.7H2O is 277.9    mass of crystals = 0.150 x 277.9  = 41.7g`

### Titrations

Titrations are one of the more common examples of reacting quantities calculations involving concentrations. A known volume of an acid or alkali with unknown concentration is exactly neutralised using a standard solution of an alkali or acid, using an indicator to see when neutralisation has been achieved. Titration can be very accurate and precise if a suitable procedure is followed to minimise sources of error.

It is normal practice to perform sufficient repeat titrations to obtain concordant results: two results within 0.1cm3 of each other. A mean of these two results is then calculated as the average titre, which is used in calculations.

### Application

A solution of HCl of unknown concentration was titrated against 25.0cm3 of KOH solution of concentration 0.500 mol dm-3. The average titre at the endpoint was 18.50cm3. Calculate the concentration of the acid.

```Firstly we need the equation for the reaction:   HCl + KOH → KCl + H2O
Moles of KOH = conc of KOH x vol of KOH (in dm3) = 0.500 x (0.0250) = 0.0125 mol
Moles of HCl = 0.0125 mol (1:1 mole ratio)
Conc of HCl = moles of HCl ÷ vol of HCl (in dm3) = 0.0125 ÷ 0.0185
= 0.676 mol dm-3 (to 3 sf)
```

### %Yield Calculations

The % yield is used to compare how much product we actually made (the yield) to how much we could have made in theory (a mole calculation).

% yield = (mass of product obtained ÷ theoretical mass of product) × 100

or     (moles of product obtained  ÷ theoretical moles of product) × 100

### Application

Methane can react with bromine when exposed to uv light, forming bromomethane. CH4(g) + Br2(g) → CH3Br(g) + HBr(g)

If 1.6g of methane reacted with excess bromine and 6.0g of bromomethane were formed, what was the % yield?

```Step 1: find out how many moles of CH3Br could be made if all CH4 reacted.
Moles of CH4 = mass ÷ Mr  = 1.6 ÷ 16 =  0.1 moles
Mole ratio from equation is 1:1 so 0.1 moles ofCH4 makes 0.1 moles of CH3Br
Mass of CH3Br = moles x Mr  =   0.1 × (12 + 3 + 79.9) = 9.49g```
```Step 2: work out % yield
% yield = (6.0  ÷ 9.49) × 100   = 63.22%```