Edexcel A-Level Chemistry AS 1 Formulae Equations and Amount of Substance Questions

Write the chemical relationship first, convert every quantity to moles with consistent units and use formula or equation ratios before reporting the requested value.

Exam points

Calculate moles, particles, masses or gas volumes using molar quantities and constants.
Write balanced full or ionic equations with correct formulae, charges and state symbols.
Apply ppm, concentration or composition data to determine a chemical amount or formula.

Covered syllabus points

1.1—The terms ‘atom', 'element', 'ion', 'molecule', 'compound', 'empirical formula' and 'molecular formula’1.2—The mole (mol) is the unit for the amount of a substance 1.3—Write balanced full and ionic equations 1.4—The terms: i ‘relative atomic mass’ based on the 12C scale ii ‘relative molecular mass’ and ‘relative formula mass’1.5—The concentration of a solution in mol dm−3 and g dm−3 Titration calculations are not required at this stage 1.6—Experimental data to calculate empirical and molecular formulae 1.7—Chemical equations to calculate reacting masses and vice versa 1.8—Chemical equations to calculate volumes of gases and vice versa 1.9—Percentage yields and percentage atom economies (by mass) in laboratory and industrial processes 1.10—Determine a formula or confirm an equation by experiment 1.11—CORE PRACTICAL 1 Measurement of the molar volume of a gas 1.12—Relate ionic and full equations, with state symbols, to observations from simple test-tube experiments, to include: i

[Maximum number: 6]

Terpenes and terpenoids are groups of chemicals found in plants.
They contribute to the smell and taste of the plant, and so are used in flavourings and perfumes.

Both terpenes and terpenoids have carbon skeletons based on one or more units of the compound isoprene.

Figure for Question 1 — Edexcel A-Level Chemistry AS

isoprene
Terpenes are made up of two or more isoprene units joined together, while terpenoids have additional functional groups, often containing oxygen.

Limonene and myrcene are both terpenes composed of two isoprene units.
Menthol, linalool and citronellol are terpenoids containing an - OH group.

(a)

Show that myrcene and limonene are terpenes by comparing the molecular and/or empirical formulae of myrcene, limonene and isoprene.

myrcene

limonene

isoprene

[ 2 ]

An answer that makes reference to the following points:

- molecular formula of either myrcene, limonene or both is $\mathrm{C}_{10} \mathrm{H}_{16}$ and the molecular / empirical formula of isoprene is $\mathrm{C}_{5} \mathrm{H}_{8}$

- so the molecular formula of myrcene and limonene is twice isoprene (so they are terpenes)

OR

- molecular formula of either myrcene, limonene or both is $\mathrm{C}_{10} \mathrm{H}_{16}$ and the molecular / empirical formula of isoprene is $\mathrm{C}_{5} \mathrm{H}_{8}$

- so the empirical formula of myrcene and limonene is the same as isoprene (so they are terpenes)

(b)

5 g samples of myrcene and limonene were completely hydrogenated. Calculate the volume of hydrogen required to hydrogenate each sample.

Include the unit with your answer.

Assume the volume of 1 mol of hydrogen $=24000 \mathrm{~cm}^{3}$

[ 4 ]

Stand alone first mark

- calculation of $M_{\mathrm{r}}$ of terpene and moles of terpene

(1)

Method 1

- either value for number of moles of $\mathrm{H}_{2}$

(1)

- either volume of $\mathrm{H}_{2}$

(1)

- second volume of $\mathrm{H}_{2}$ with unit

(1)

Method 2

- either value for one $\mathrm{H}_{2}$ per terpene

(1)

- either volume of $\mathrm{H}_{2}$

- second volume of $\mathrm{H}_{2}$ with unit

(1)

\multirow{4{*}{Example of calculation:

$5 \div 136=0.036765$

& Myrcene & Limonene

numbers of moles of

hydrogen required

$0.036765 \times 3$

=0.11030

$0.036765 \times 2$

=0.07353

volume of hydrogen

required

$0.11030 \times 24000$ $=2647.2 \mathrm{~cm}^{3} / 2.6472 \mathrm{dm}^{3}$ $0.07353 \times 24000$ $=1764.7 \mathrm{~cm}^{3} / 1.7647 \mathrm{dm}^{3}$

\end{tabular}
\end{tabular}}}

(4)

\multirow{3{*}{& Myrcene & Limonene

one hydrogen per

terpene

$0.036765 \times 24000$

=882.36

$0.036765 \times 24000$

=882.36

total volume of

hydrogen required

$882.36 \times 3$ $=2647.1 \mathrm{~cm}^{3} / 2.6471 \mathrm{dm}^{3}$ $882.36 \times 2$ $=1764.7 \mathrm{~cm}^{3} / 1.7647 \mathrm{dm}^{3}$

\end{tabular}}}

Final answers can be found by use of ratios of 2:3 or 3:2 so not all values will necessarily be seen

Ignore rounding errors in all steps except the final answer

Both final volumes attributed to the correct terpene with some working scores (4)

Allow TE throughout

Ignore SF

[Maximum number: 9]

This question is about ethene, $\mathrm{C}_{2} \mathrm{H}_{4}$ .

(a)

The global production of ethene is about 150 million tonnes per year.

Calculate the number of molecules in 150 million tonnes of ethene.
$\left[1\right.$ tonne $=1000 \mathrm{~kg} \quad$ Avogadro constant $\left.=6.02 \times 10^{23} \mathrm{~mol}^{-1}\right]$

[ 2 ]

(b)

Ethene is used commercially to speed up the ripening of bananas.

Ethene levels of 100 parts per million (ppm) are used in ripening rooms.
Calculate the amount, in moles, of ethene in a ripening room of volume $220 \mathrm{~m}^{3}$ at a temperature of $21^{\circ} \mathrm{C}$ and pressure of $1.01 \times 10^{5} \mathrm{~Pa}$ .

Give your answer to an appropriate number of significant figures.

\left[p V=\mathrm{n} R T \quad R=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right]

[ 5 ]

(c)

Ethene reacts with oxygen, $\mathrm{O}_{2}$ , in the presence of a silver catalyst, to form compound Y as the only product.

[ 2 ]

(i)

When 10.0 g of ethene reacts completely with oxygen, the mass of compound Y formed is 15.7 g.

Calculate the empirical formula of Y.
You must show your working.
[A, values: H=1.0, C=12.0, O=16.0 ]

[ 2 ]

Method 1 - calculation of mass C, H and O (1) - calculation of moles C, H and O and Ignore $\ce{SF}$ except 1 $\ce{SF}$ for all methods Example of calculation: mass $\ce{C=24}$ /28 x 10.0=8.5714( g) mass $\ce{H=4}$ /28 x 10.0=1.4286( g) mass $\ce{O=15.7-}$ 10.0=5.70( g)arrayc:c:c C &: & H 8.5714/12 &: & 1.4286/1 array: $\ce{O5.7}$ empirical formula empirical formula is C_2 H_4 O Method 2 - calculation of moles C_2 H_4 and O 1 calculation of moles C_2 H_4 and O_2 (moles C_2 H_4=10.0/28= ) 0.35714 (. moles .O_2=(15.7-10.0)/32=) 0.17813 - empirical formula M2 dependent on M1 empirical formula is C_2 H_4 O Method 3 - calculation of moles C_2 H_4 and (moles .C_2 H_4=) 10.0/28=0.35714 (2) M_r product (1) (M_r. product .=) 15.7/0.35714=43.96 - empirical formula and calculation of M_r of empirical formula (1) M2 dependent on M1 empirical formula is C_2 H_4 $\ce{O2}$ x 12+4 x 1+1 x 16=44

[Maximum number: 9]

Magnesium carbonate powder reacts with hydrochloric acid.

(a)

Complete the equation for this reaction by adding state symbols.

[ 1 ]

all correct state symbols

$\mathrm{MgCO}_{3}(\mathbf{s})+2 \mathrm{HCl}(\mathbf{a q}) \rightarrow \mathrm{MgCl}_{2}(\mathbf{a q})+\mathrm{H}_{2} \mathrm{O}(\mathbf{l})+\mathrm{CO}_{2}(\mathbf{g})$

Allow capital letters Ignore extra brackets

(1)

Figure for Answer to Question (a) — Edexcel A-Level Chemistry AS

(b)

A student carried out an investigation to determine the molar volume of carbon dioxide using this apparatus.

Figure for Question (b) — Edexcel A-Level Chemistry AS

The student carried out five experiments, adding a different mass of magnesium carbonate each time. The results are shown in the table.

Table for Question (b) — Edexcel A-Level Chemistry AS

[ 6 ]

(i)

Calculate the molar volume of carbon dioxide using your answers to parts (b)(ii) and (b)(iii). Give your value to an appropriate number of significant figures and include units.

\left[\begin{array}{lll} A_{\mathrm{r}} \text { values: } & \mathrm{Mg}=24.3 & \mathrm{C}=12.0 \end{array} \mathrm{O}=16.0\right]

[ 4 ]

calculation of molar mass of magnesium carbonate (1)

- calculation of moles of magnesium carbonate

(1)

- calculation of molar volume

(1)

- answer given to 2 or 3 SF and units (1) M4 dependent on award of M3

(4)

Example of calculation:

84.3

OR expression used correctly:
$[24.3+12+(3 \times 16)]$ $46 \div 0.0021352=21543 / 2.1543 \times 10^{4}\left(\mathrm{~cm}^{3}\right)$ $=22 \mathrm{dm}^{3}\left(\mathrm{~mol}^{-1}\right) / 22000 \mathrm{~cm}^{3}\left(\mathrm{~mol}^{-1}\right)$
Or $21.5 \mathrm{dm}^{3}\left(\mathrm{~mol}^{-1}\right) / 21500 \mathrm{~cm}^{3}\left(\mathrm{~mol}^{-1}\right)$
TE on any reasonable pair of values obtained
from the candidates' graph or table provided
eg $54 \mathrm{~cm}^{3}$ and $0.215(\mathrm{~g}) \rightarrow 2.5504 \times 10^{-3}(\mathrm{~mol}) \rightarrow 21$ $200 \mathrm{~cm}^{3}$

Correct answer scores 4 marks

Final answer must not be given as a fraction to get MP4

Ignore units except for MP4

(ii)

The acid must be in excess for each experiment. Calculate the minimum concentration of hydrochloric acid needed for $30 \mathrm{~cm}^{3}$ of acid to completely react with 0.25 g of magnesium carbonate.

\mathrm{MgCO}_{3}+2 \mathrm{HCl}+\mathrm{MgCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}

[ 2 ]

moles of magnesium carbonate and moles of acid in $30 \mathrm{~cm}^{3}$

(1)

- calculation of minimum concentration with units (1)

Example of calculation:
$\begin{aligned} & \mathrm{n}=0.25 / 84.3 & \mathrm{n}=0.0029655 \text { or } 0.00297 \end{aligned}$

and

1:2 stoichiometry

$\therefore 0.00593$ (moles acid)

Accept 0.00594 from 0.00297

$(0.00593 / 30) \times 1000=0.198 \mathrm{~mol} \mathrm{dm}^{-3}$

Accept answers from

0.198 to $0.200 \mathrm{~mol} \mathrm{dm}^{-3}$

Allow TE throughout e.g. $M_{\mathrm{r}}$ from 21(b)(iv) Ignore SF

Correct answer with no working scores 2

(2)

(c)

The value of molar volume calculated in (b)(iv) was lower than the student expected. Give two reasons for the value being lower than expected. Assume that the correct amounts of hydrochloric acid and magnesium carbonate were used. (2)

[ 2 ]

An answer that makes reference to any two of the following points:

- loss of gas before the bung is inserted / other named reason

(1)

- some carbon dioxide dissolved in the water

(1)

- temperature of the lab was lower than standard temperature.

(1)

Do not allow "loss of gas" unless a reason is given eg delivery tube not positioned correctly so not all goes into measuring cylinder, badly fitting bung Ignore leaks

Allow gas for carbon dioxide

Ignore higher pressure

Do not award higher temperature / lower pressure / suck-back

Ignore impurities in $\mathrm{MgCO}_{3}$

Ignore incomplete reaction

Comment: Apply the list principle ie

- 1 correct answer and 1 do not award answer scores 1

- 2 correct answers and 1 do not award answer scores 1

- 2 correct answers and 2 do not award answers scores 0

- 2 correct answers and 1 ignore scores 2

(2)

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question (c)

Question (i)

Question 3

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)