Question

What is the percent by mass of oxygen in a gaseous mixture whose molar composition is 0.500 % CO2 and 99.500% air? The composition of air is 21% mole O2, 79 % mole N2 , and has an average molar mass of 29.0 g/mol Number % O by mass

Answer 1

23.0 %

Step-by-step explanation:

Let the gaseous mixture has total 100 moles

Composition of
`CO_2` = 0.500 %

The moles of
`CO_2` =
`\frac {0.500}{100}* 100\ moles` = 0.5 moles

Also,

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`0.5\ moles= (Mass)/(44.01\ g/mol)`

Mass of
`CO_2` = 22.005 g

Composition of air = 99.500 %

The moles of air =
`\frac {99.500}{100}* 100\ moles` = 99.500 moles

Also,

Given, Average molar mass of air = 29.0 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`99.5\ moles= (Mass)/(29.0\ g/mol)`

Mass of air = 2885.5 g

Total mass of 100 moles of the mixture = 22.005 g + 2885.5 g = 2907.505 g

Composition of
`O_2` = 21 % of air

The moles of
`O_2` =
`\frac {21}{100}* 99.5\ moles` = 20.895 moles

Also,

Molar mass of oxygen = 31.9988 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`20.895\ moles= (Mass)/( 31.9988\ g/mol)`

Mass of
`O_2` = 668.615 g

Thus,

% O by mass =
`\frac {Mass\ of\ Oxygen}{Total\ mass}* 100=\frac {668.615}{2907.505}* 100` = 23.0 %