 # 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 O2 not just O.

The real question is: How many 6 x 1023 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

Example 2

Calculate the percent by mass of each element of the compound methanol CH4O.

We recommend that you tackle this problem by:

Comparing the weights of each element within CH4O.

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

H4 = There are 4 protons in hydrogen4

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 H4 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 H4 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%

Hydrogen4 as a % =

 100% = 32 grams x% 4 grams 100% = 32 x% 4

Rearranging the formula:

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

H4 = 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 CH4O

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 CH4O

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 C2H6O or methanol CH4O?

Test C2H6O for comparison of % of mass in each element.

Comparing the weights of each element C2H6O.

C = There are 12 protons and neutrons in carbon

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

C2 = There are 24 protons and neutrons in C

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

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

H6 = There are 6 protons in hydrogen6

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:

 C2 to H6 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:

 C2 to H6 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%

C2H6O fulfils the requirements as the mystery liquid.

To double check we can check CH4O using the same method.

Comparing the weights of each element of CH4O.

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

H4 = There are 4 protons in hydrogen4

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 H4 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 H4 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  C2H6O 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 C2H6O or methanol CH4O?

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 C2H6O contains 2 mol of carbon, 6 mol of hydrogen and 1 mol of oxygen. So the molar mass of C2H6O is (12x2)+(1x6)+(16x1)= 46g

While CH4O contains 1 mol of carbon, 4 mol of hydrogen and 1 mol of oxygen. So the molar mass of CH4O 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 C2H6O

 %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 CH4O

 %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 C2H6O

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.