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# AS Chemistry Revision - Calculations and Exam Questions

## Equations

__Moles of a solid compound:__

Moles = Mass ÷ Relative Formula Mass

__Moles of a solution:__

Moles = (Volume x Concentration) ÷ 1000

__Moles of a gas:__

Moles = Volume of gas ÷ Volume of 1 mole (24,000cm^{3})

__Amount of atoms:__

Amount of atoms = Moles x Avogadro's constant (6.02 x 10^{23})

## Step-By-Step Exam Questions

__Question 1 - Part 1 (2 marks):__

**'Calcium carbonate,** CaCO

_{3}

**,**reacts with

**nitric acid,**HNO

_{3}.

*A student neutralised 2.68g of CaCO_{3 }with 2.50 mol dm^{-3} nitric acid, HNO_{3}.*

*The equation for this reaction is shown below:*

**CaCO _{3(s) }+ 2HNO**

_{3}

_{(a}

_{q)}**-> Ca(NO**

_{3}**)**

_{2}

_{(aq)}**+ CO**

_{2}**+ H**

_{2}**O**

*Determine the amount, in mol, of CaCO_{3 }reacted.'*

- Using the formula above for calculating the moles of a solid compound, we need to
**divide**the**mass**by the molecular formula mass**(Mr).**

- Add up the atomic masses of
**Ca**(40.1),**C**(12) and**3 O's**(16 x 3):

40.1 + 12 +16 + 16 + 16 = 100.1

- Now
**divide**the**mass**of CaCO_{3}by the**Mr**that you just calculated:

2.68 ÷ 100.1 = 0.0268.

- The amount of CaCO
_{3 }reacted was__0.0268 mol.__

__Question 1 - Part 2 (1 mark):__

*'Now calculate the volume, in cm*

^{3}*, of*

*CO*_{2 }*produced at room temperature and pressure*

**(RTP)**.'- First of all look for the equation that involves the volume of a gas:
**Moles = Volume of gas****÷ 24,000cm**^{3}**.**

- We now have to
**rearrange**the equation in order to find the volume of gas. We already know the amount of moles as we worked it out in the question above. The amount of**moles**in a compound is the**same**for all the different**compounds**in a reaction. If you work out the amount of moles of CaCO3, the amount of moles of CO_{2 }will be the same.

- So, to rearrange the equation you take the
**'****÷ 24,000'**part and put it on the other side of the**=**, changing the**÷**sign to a**x**sign.

- You then have the equation
**Moles x 24,000 = Volume of gas.**

- Now substitute the figures that you know into the equation:
**0.0268 x 24,000 = 640.8**

- Therefore the amount of CO
_{2 }produced in the reaction of calcium carbonate and nitric acid is**643.2cm**^{3}.

__Question 1 - Part 3 (2 marks):__

*'Now calculate the volume of 2.50 mol dm*

^{-3}

**HNO**

_{3}*needed to neutralise 2.68g of*

**CaCO**

_{3.}- First of all find the equation that involves the volume of a solution:

Moles = (Volume x Concentration) ÷ 1000

- Now you need to
**rearrange**the equation to make the volume the thing we are trying to find:**(Moles x 1000) ÷ Concentration = Volume.**

- If you look at the original equation:

CaCO_{3(s) }+ 2HNO_{3}_{(a}_{q)}-> Ca(NO_{3})_{2}_{(aq)}+ CO_{2}+ H_{2}O

you can see that there are**2**lots of the compound HNO_{3. }This means that we have to**double**the amount of**moles**that we calculated for CaCO_{3}, so:

0.0268 mol x 2 =**0.0536 mol.** - Now
**substitute**all of the numbers you know into the equation (Moles x 1000) ÷ Concentration = Volume:

(0.0536 x 1000) ÷ 2.50 = 21.44

- Therefore the volume of HNO
_{3}needed to neutralise 2.68g of CaCO_{3}is**21.44cm**^{3}.

Question 2 - Part 1 (1 Mark):

**'Graphite** is used in pencils but is still referred to as

**'pencil lead'.**

*A student decided to investigate the*

**number of carbon atoms**in a 'pencil lead'. He found that the 'pencil lead' was**0.321g.****Calculate** the amount, in **mol**, of carbon **atoms** in the student's pencil lead (assume that the 'pencil lead' is **pure graphite**).'

- Much like question 1, we must use the equation for working out the amount of moles of a solid:
**Moles = Mass ÷ Mr (relative formula mass)**

- Graphite is a form of
**carbon**and therefore you use the**Mr**of carbon**(12)**in the equation.

**Substitute**the values into the equation:

Moles = 0.321 ÷ 12 = 0.0268

- Therefore the amount of carbon atoms, in mol, in the student's 'pencil lead' is
**0.0268 mol.**

Question 2 - Part 2 (1 Mark):

*'Using the Avogadro's constant calculate the number of carbon atoms in the student's 'pencil lead'.'*

- Using the equation
**amount of atoms = moles x Avogadro's constant**we can literally**substitute**in the values and find the answer!

- We worked out the moles in the previous part of the question so we know that the moles of the 'pencil lead' is
**0.0268:**

0.0268 x 6.02 x 10^{23}= 1.61x10^{22}

- Therefore, we can conclude that the number of carbon atoms in the student's 'pencil lead' is
**1.61x10**^{22}**.**

All of these questions were found from past papers on the OCR website.

If you found this hub useful you might want to read this hub to learn the basic concepts you'll need to know for your exam.