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How do we calculate concentration, titration results and percentage yield?

Concentration in g per dm cubed and mol per dm cubed, using titration results to find an unknown concentration, and calculating percentage yield.

A CCEA GCSE Chemistry answer on further calculations, covering concentration in g per dm cubed and mol per dm cubed, how to use titration results and a balanced equation to find an unknown concentration, and how to calculate percentage yield.

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  1. What this dot point is asking
  2. Concentration
  3. Titration calculations
  4. Percentage yield
  5. Worked example
  6. Examples in context
  7. The titration practical
  8. Why yields are below 100 percent
  9. Try this

What this dot point is asking

CCEA wants you to calculate concentration in g per dm cubed and mol per dm cubed, use titration results with a balanced equation to find an unknown concentration, and calculate percentage yield.

Concentration

To convert between the two, use moles equals mass divided by relative formula mass. The most common slip is forgetting to change cm cubed into dm cubed, so always do that conversion first.

Titration calculations

Percentage yield

Yields are usually less than 100 percent because some product is lost in handling, the reaction may not be complete, or side reactions occur. A high percentage yield means an efficient process.

Worked example

Examples in context

Example 1. Checking the strength of vinegar
Food scientists titrate vinegar against a standard alkali to find its concentration of ethanoic acid, ensuring it meets the labelled strength. The same titration method used in the lab is a routine quality-control test.
Example 2. Improving an industrial yield
A manufacturer calculates the percentage yield of a process and works to raise it, because a higher yield means less wasted raw material and lower costs. The yield calculation links chemistry directly to economic efficiency.
Example 3. Dosing a swimming pool
Pool operators work out how much chlorine compound to add per cubic metre of water to reach a safe concentration. The same concentration calculation used in the lab, amount of substance per unit volume, scales up to keeping pools and drinking water safe.

The titration practical

The titration calculation rests on a careful practical. A known volume of one solution is measured into a conical flask with a pipette, an indicator is added, and the other solution is run in from a burette until the colour just changes at the end point. The burette reading gives the volume that reacted. Good technique matters: swirl the flask, add dropwise near the end point, and repeat until two readings agree closely (concordant results), then average them. CCEA often gives a marks for describing this method as well as for the calculation, so learn the apparatus (pipette, burette, conical flask) and the steps alongside the sums.

Why yields are below 100 percent

It is worth being able to explain a low yield, not just calculate it. Product can be lost when transferring between containers, the reaction may not go to completion, or competing side reactions may use up some reactant. A reversible reaction, such as the one in the Haber process, can never reach 100 percent yield because it settles at equilibrium. Recognising these reasons turns a yield figure into a piece of chemical reasoning, which is exactly what higher-mark questions reward.

Try this

Q1. State the formula for percentage yield. [1 mark]

  • Cue. Actual yield divided by theoretical yield, times 100.

Q2. Convert 50 cm cubed into dm cubed. [1 mark]

  • Cue. 50 divided by 1000 equals 0.050 dm cubed.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20194 marksIn a titration, 25.0 cm3 of sodium hydroxide solution is neutralised by 20.0 cm3 of 0.10 mol/dm3 hydrochloric acid. The equation is NaOH + HCl gives NaCl + H2O. Calculate the concentration of the sodium hydroxide in mol/dm3.
Show worked answer →

Markers want moles of acid, the 1:1 ratio, then the concentration.

Moles of HCl = concentration times volume in dm3 = 0.10×20.01000=0.00200.10 \times \frac{20.0}{1000} = 0.0020 mol.

The equation is 1:1, so moles of NaOH = 0.0020 mol.

Concentration of NaOH = moles divided by volume in dm3:

c=0.002025.0/1000=0.00200.0250=0.080 mol/dm3c = \frac{0.0020}{25.0/1000} = \frac{0.0020}{0.0250} = 0.080 \text{ mol/dm}^3

Markers reward moles of acid (0.0020), the 1:1 ratio giving moles of NaOH, and the concentration 0.080 mol/dm3.

CCEA 20213 marksA reaction has a theoretical (maximum) yield of 8.0 g of product, but only 6.0 g is actually obtained. Calculate the percentage yield.
Show worked answer →

The marks are for the formula and the answer.

Percentage yield is the actual mass obtained divided by the theoretical maximum mass, times 100:

% yield=actual yieldtheoretical yield×100=6.08.0×100\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{6.0}{8.0} \times 100

Working it out: 6.0÷8.0=0.756.0 \div 8.0 = 0.75, then ×100=75%\times 100 = 75\%.

Markers reward the percentage yield formula, correct substitution, and the answer 75 percent.

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