# Amount of Substance

All you need to know for AQA AS Chemistry, Amount Of Substance topic :) A grade revision cards :)

- Created by: Natalie Beard
- Created on: 15-12-09 15:27

## Moles and Avogadro's Constant

Avogadro

- 1 mole of any substance contains the Avogadro number of particles
- A particle can be a molecule, ion, electron, proton and neutron.
- Equal moles of substances contain equal numbers of particles
- The Avogadro Number is 6.023 x 10(23)

Avogadro Calculation

**Ar or Mr = A.C x mass on one atom/molecule**

Moles: The number of moles in an substance can be calculated using this formula:

**n (Number of moles) = mass (g) / Mr/Ar**

**Aka a formula triangle with Mass at the top and others below**

## Ideal Gas Equation

**PV= nRT**

P= Pressure in Pascals (Pa)

n= Number of moles of gas

R = Gas constant: 8.31JK(-1)mol(-1)

T= Temperature in Kelvin (K

V=Volume in m(3)

## Empirical Formula

**The simplest whole number ratio of the number of atoms of each element present in a molecule**

Best explained by an example First divide each value by the atomic mass of each.

**1. 2.4% H = 2.4/1= 2.4
39.1% S = 39.1/32 = 1.222
58.55O = 58.5/16 = 3.66
2. Divide each mol value by the smallest value
H = 2.4/1.2 = 2
S = 1.2/1.2 = 1
O = 3.66/1.2 = 3**

3. Empirical formula is **H2SO3**

## Molecular Formula

**The molecular formula shows the actual number of atoms of each element present in a molecule of a compound**

Again, best shown by example, using the question on the previous card:

1. **Add up the empirical mass**: H(2)SO(3) = (1 x 2) + 32 + (16x3)= 82

2. **Identify the Mr**, usually provided in question, or you had to calculate it for previous question, in this case: 164.2

3. 164.2/82 = 2 = this is the **ratio between empirical and molecular formula**

4. **Times all the elements in empirical formula by the ratio**, in this case, 2: so H(2)SO(3) becomes H(4)S(2)O(6)

## Reacting Volumes of Gas

**1 mole of any gas at room temperature, (25 degrees) and standard pressure, (100kPa) occupies a volume of 24dm(3)****Equal volumes of gases contain equal numbers of moles: 1 mole of H(2) occupies a volumne of 24dm(3), as does CO(2) and CH(4)****In a gaseous reaction the ratio of the volumes of gases is the same as the mole ratio of the gaseous reactants and products:**

Example: Sulphur dioxide and oxygen are required to form sulphur trioxide according to equation:

2SO(2) + O(2) --------------------> 2SO(3)

a. Calculate the volume of O(2) needed to completely react with 66cm(3) of SO(2)?

b. Volume of S0(3) produced from 66cm(3) of SO(2)?

c. Volume of O(2) that remains unreacted if 70cm(3) are added to 100cm(3) of SO(2)?**Answers on reverse**

## Reacting Volume of Gas (Answers)

a. Mole ratio SO(2) to O(2) is 2:1 so **33cm(3)m of O(2)**

b. Mole ratio SO(2) to SO(3) is 1:1 so **66cm(3) SO(3)**

c. Mole ratio SO(2) to O(2) is 2:1 so 50cm(3) of O(2) are needed to react with 100cm(3) of SO(2), leaving **20cm(3) unreacted**.

## Concentration in Solution

- The number of moles of solute in 1dm(3) of solution, units are
**moldm(-3)**

- A solution containing 1 mole of a solute in 1 dm(3) of solution is said to have a concentration of 1moldm(-3)

- Also, a 0.5moldm(-3) solution will have 0.5 moles of solute in 1 dm(3) of solution

To calculate the number of moles in a solute in a given volume of solution a simple formula is used:

**Moles = Concentration x Volume/1000**

(if volume is already in **dm(3)** **no need to divide by 1000**)

## Atom Economy

**A measure of how much of a desired product in a reaction is formed from the reactants. It is a theoretical quantity calculated from a balanced equation**

Formula

**Mass of desired products/total mass of reactants x 100**

Example

CH(4) + 2Cl(2) ----------------------> CH(2)Cl(2) + 2HCL

A.E= Mass of 1 mole of CH(2)Cl(2)/mass of one mole of CH(4) + 2 moles of Cl(2) x 100

So= 85/16+ 42 x 100 = 53.8% is the desired product

This is essentially a measure of how useful a particular reaction is for achieving a particular product

## Percentage Yield

**A practical measure of the efficiency of a reaction**.

It takes into account reactions that do not go to completion, can only be calculated from experimental data.

**Actual mass of product/maximum theoretical mass of product x 100**

Example

In an experiment 21.3g of CH(2)Cl(2) were produced when 8.0g of methane were reacted with an excess of chlorine, what is the percentage yield?

CH(4) + 2Cl(2) ---------------------> CH(2)Cl(2) + 2HCL

**Answer on reverse**

## Percentage Yield (Answer)

n of CH(4) = Mass/Mr = 8.0/16.0 = 0.50 mol

Maximum n of CH(2)Cl(2) that can be formed from 0.50 mol of CH(4) = 0.50 mol

Maximum theoretical mass = n x Mr = 0.50 x 85.0 = 42.5g

Actual mass of CH(2)Cl(2) formed = 21.3g

% yield: 21.3/42.5 x 100 = 50.1%

This suggests reaction didn't go to completion or that some of the CH(4) was converted into by-products

## Conversions for Ideal Gas Equation

**Pressure must be in Pa so:**

If given in kPa: x 1000 = Pa

If given in mPa x 1000000 = Pa

**Volume must be in m(3)**

If given in dm(3) divide by 1000 = m(3)

If given in cm(3) divide by 1000000 = m(3)

**Temperature must be in Kelvin (K)**

If given in degrees add 273 = K

## Comments