# Examples

**Example 1**

Calculate the number of moles of oxygen molecules in a sample weighing 8 grams.

**Note: **

This is a trick.

Because this says *molecules* and not *atoms* of oxygen you know that molecules are at least 2 atoms. So they are looking for O_{2} __not__ just O.

The real question is: How many 6 x 10^{23} molecules of oxygen (in grams) go into a sample weighing 8 grams?

How many protons and neutrons are there in 1 molecule of oxygen = 2 x 16 = 32

1 proton or neutron weighs \frac{1}{6\times10^{23}}

Therefore 32 protons and neutrons weigh 32 x \frac{1}{6\times10^{23}} grams

One mole of these = 6\times10^{23}\times\frac{32}{6\times10^{23}} = 32 grams

Therefore the sample of 8 grams of oxygen is 0.25 of this

**Answer = 0.25 moles**

**Example 2**

Calculate the percent by mass of each element of the compound methanol CH_{4}O.

We recommend that you tackle this problem by:

Comparing the weights of each element within CH_{4}O.

C = There are 12 protons and neutrons in carbon

each weighing \frac{1}{6\times10^{23}} grams

H = There is 1 proton weighing \frac{1}{6\times10^{23}} grams

H_{4} = There are 4 protons in hydrogen_{4}

each weighing \frac{1}{6\times10^{23}} grams

O = There are 16 protons and neutrons in oxygen

each weighing \frac{1}{6\times10^{23}} grams

So we are comparing:

C |
to | H_{4} |
to | O |
= | Total |

12\times\frac{1}{6\times10^{23}} | 4\times\frac{1}{6\times10^{23}} | 16\times\frac{1}{6\times10^{23}} | 32\times\frac{1}{6\times10^{23}} |

Because \frac{1}{6\times10^{23}} are in each of them, we are really comparing:

C |
to | H_{4} |
to | O |
= | Total |

12 | 4 | 16 | 32 |

i.e.

ratio 12:4:16

Then using maths:

**Carbon** as a % =

100% | = | 32 | grams |

x% | 12 | grams | |

100% | = | 32 | |

x% | 12 |

Rearranging the formula:

x% | = | \frac{100\times12}{32} | = | 37.5% |

C=37.5%

**Hydrogen _{4}** as a % =

100% | = | 32 | grams |

x% | 4 | grams | |

100% | = | 32 | |

x% | 4 |

Rearranging the formula:

x% | = | \frac{100\times4}{32} | = | 12.5% |

H_{4} = 12.5%

**Oxygen** as a % =

100% | = | 32 | grams |

x% | 16 | grams | |

100% | = | 32 | |

x% | 16 |

Rearranging the formula:

x% | = | \frac{100\times16}{32} | = | 50% |

O = 50%

% Comparison by mass:

Carbon = 37.5%

Hydrogen = 12.5%

Oxygen = 50%

Below is how many schools teach how to tackle the problem __but__ we DO NOT advise doing it this way because our way is so much easier.

Here goes:

Calculate the percent by mass of each element of the compound methanol CH_{4}O

**Answer:**

**1. Write out what you know**

1 mole of carbon | = | \strike{6\times10^{\strike{23}}\times\frac{12}{\strike{6\times10^{\strike{23}}}}} | = | 12g |

1 mole of hydrogen | = | \strike{6\times10^{\strike{23}}\times\frac{1}{\strike{6\times10^{\strike{23}}}}} | = | 1g |

1 mole of oxygen | = | \strike{6\times10^{\strike{23}}\times\frac{16}{\strike{6\times10^{\strike{23}}}}} | = | 16g |

One mole of the compound contains 1 mol of carbon, 4 mol of hydrogen and 1 mol of oxygen.

The compound mass is therefore (12x1)+(1x4)+(16x1) = 32g

So there is 32g/mol CH_{4}O

100% = 32g

Ratio by mass of carbon, hydrogen and oxygen within the compound is:

Carbon (12 x 1) : Hydrogen (1 x 4) : Oxygen (16 x 1) = 12:4:16 = 3:1:4

**2. Calculate the percentage mass of each element within the compound**

As one mole of carbon has a mass of 12g and the total mass of the entire compound is 32g the percentage of carbon is:

%C | = | \frac{12\times1}{32}\times100 | = | 37.5% |

One mole of hydrogen has a mass of 1g and there are 4 moles of hydrogen so the percentage of hydrogen is:

%H | = | \frac{1\times4}{32}\times100 | = | 12.5% |

One mole of oxygen has a mass of 16g, so the percentage of hydrogen is:

%O | = | \frac{16\times1}{32}\times100 | = | 50% |

**Example 3**

After distilling a fermented solution in the lab the resulting liquid is found to contain more than 50% carbon and less than 35% oxygen by mass. Is the liquid ethanol C_{2}H_{6}O or methanol CH_{4}O?

Test C_{2}H_{6}O for comparison of % of mass in each element.

Comparing the weights of each element C_{2}H_{6}O.

C = There are 12 protons and neutrons in carbon

each weighing \frac{1}{6\times10^{23}} grams

C_{2} = There are 24 protons and neutrons in C_{2 }

each weighing \frac{1}{6\times10^{23}} grams

H = There is 1 proton weighing \frac{1}{6\times10^{23}} grams

H_{6} = There are 6 protons in hydrogen_{6}

each weighing \frac{1}{6\times10^{23}} grams

O = There are 16 protons and neutrons in oxygen

each weighing \frac{1}{6\times10^{23}} grams

So we are comparing:

C_{2} |
to | H_{6} |
to | O |
= | Total |

24\times\frac{1}{6\times10^{23}} | 6\times\frac{1}{6\times10^{23}} | 16\times\frac{1}{6\times10^{23}} | 46\times\frac{1}{6\times10^{23}} |

Because \frac{1}{6\times10^{23}} are in each of them, we are really comparing:

C_{2} |
to | H_{6} |
to | O |
= | Total |

24 | 6 | 16 | 46 |

i.e. ratio 24:6:16

Then using maths:

**Carbon** as a % =

100% | = | 46 | grams |

x% | 24 | grams | |

100% | = | 46 | |

x% | 24 |

Rearranging the formula:

x% | = | \frac{100\times24}{46} | = | 52.17% |

C=52.17%%

**Oxygen** as a % =

100% | = | 46 | grams |

x% | 16 | grams | |

100% | = | 46 | |

x% | 16 |

Rearranging the formula:

x% | = | \frac{100\times16}{46} | = | 34.78% |

O = 34.78%

__ __

C_{2}H_{6}O for fills the requirements as the mystery liquid.

To double check we can check CH_{4}O using the same method.

Comparing the weights of each element of CH_{4}O.

__ __

C = There are 12 protons and neutrons in carbon

each weighing \frac{1}{6\times10^{23}} grams

H = There is 1 proton weighing \frac{1}{6\times10^{23}} grams

H_{4} = There are 4 protons in hydrogen_{4}

each weighing \frac{1}{6\times10^{23}} grams

O = There are 16 protons and neutrons in oxygen

each weighing \frac{1}{6\times10^{23}} grams

So we are comparing:

C |
to | H_{4} |
to | O |
= | Total |

12\times\frac{1}{6\times10^{23}} | 4\times\frac{1}{6\times10^{23}} | 16\times\frac{1}{6\times10^{23}} | 32\times\frac{1}{6\times10^{23}} |

Because \frac{1}{6\times10^{23}} are in each of them, we are really comparing:

C |
to | H_{4} |
to | O |
= | Total |

12 | 4 | 16 | 32 |

i.e. ratio 12:4:16

Then using maths:

**Carbon** as a % =

100% | = | 32 | grams |

x% | 12 | grams | |

100% | = | 32 | |

x% | 12 |

Rearranging the formula:

x% | = | \frac{100\times12}{32} | = | 37.5% |

C=37.5%

**Oxygen** as a % =

100% | = | 32 | grams |

x% | 16 | grams | |

100% | = | 32 | |

x% | 16 |

Rearranging the formula:

x% | = | \frac{100\times16}{32} | = | 50% |

O = 50%

So oxygen is 50% and does not fit the criteria so again C_{2}H_{6}O is our mystery liquid.

Below again is how many schools teach how to tackle the problem but we DO NOT advise doing it this way because our way is so much easier.

Here goes:

After distilling a fermented solution in the lab the resulting liquid is found to contain more than 50% carbon and less than 35% oxygen by mass. Is the liquid ethanol C_{2}H_{6}O or methanol CH_{4}O?

**Answer:**

**1. Write out what you know**

1 mole of carbon | = | \strike{6\times10^{\strike{23}}\times\frac{12}{\strike{6\times10^{\strike{23}}}}} | = | 12g |

1 mole of hydrogen | = | \strike{6\times10^{\strike{23}}\times\frac{1}{\strike{6\times10^{\strike{23}}}}} | = | 1g |

1 mole of oxygen | = | \strike{6\times10^{\strike{23}}\times\frac{16}{\strike{6\times10^{\strike{23}}}}} | = | 16g |

Therefore C_{2}H_{6}O contains 2 mol of carbon, 6 mol of hydrogen and 1 mol of oxygen. So the molar mass of C_{2}H_{6}O is (12x2)+(1x6)+(16x1)= 46g

While CH_{4}O contains 1 mol of carbon, 4 mol of hydrogen and 1 mol of oxygen. So the molar mass of CH_{4}O is (12x1)+(1x4)+(16x1)= 32g

Carbon % of distilled solution ˃ 50%

Oxygen % of distilled solution ˂ 35%

**2. Calculate the percentage of carbon and oxygen in methanol and ethanol**

For C_{2}H_{6}O

%C | = | \frac{12\times2}{46}\times100 | = | 52.17% |

%H | = | \frac{1\times6}{46}\times100 | = | 13.04% |

%O | = | \frac{16\times1}{46}\times100 | = | 34.78% |

For CH_{4}O

%C | = | \frac{12\times1}{32}\times100 | = | 37.5% |

%H | = | \frac{1\times4}{32}\times100 | = | 12.5% |

%O | = | \frac{16\times1}{32}\times100 | = | 50% |

Therefore the liquid must be ethanol C_{2}H_{6}O

__ __

It is a good idea to also calculate the percentage of hydrogen contained within each compound as well. You can use this to check that all the percentages add up to 100% for each compound, so you can be sure you have worked it out correctly.

The percentages of carbon, hydrogen and oxygen within the methanol compound add up to 99.99% due to each percentage being rounded to two decimal figures.