A gaseous compound is found to have the following composition: 30.5% nitrogen, and the rest oxygen. the molar mass of this gas is 91.8. What are the empirical and molecular formulas of the compound

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llltkl | College Teacher | (Level 3) Valedictorian

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Percentage of nitrogen present in the compound=30.5

Hence, percentage of oxygen present in the compound=(100-30.5)

=69.5

Assume exactly 100 g of this substance. This means in 100 g of this compound, 30.5 g will be due to nitrogen and 69.5 g will be due to oxygen.

Convert each of these masses to moles, using their respective atomic weights:

30.5 g nitrogen= `30.5/14` =2.18 mol

69.5 g oxygen= `69.5/16` =4.34 mol

A stoichiometric comparison between the elements can be made to determine the empirical formula. This is achieved by dividing through each of the mole quantities by which ever mole quantity is the smallest number of moles.

Here, the smallest mole quantity is the moles of nitrogen. So,

Nitrogen=`2.18/2.18=1`

Oxygen=`4.34/2.18=1.99~~2`

The ratio of N:O=1:2

Thus, the empirical formula of the given compound is `NO_2` .

The molar mass of this compound is 91.8 g/mol

Its empirical mass=46 g/mol

Divide the molar mass by the empirical mass to find a multiple,n:

`91.8/46=1.99~~2`

The molecular formula is a multiple of 2 times the empirical formula: `(NO_2)_n` where `n=2`

=`N_((1*2))O_((2*2))=N_2O_4.`

Thus, the molecular formula of the given compound is `N_2O_4` .

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ayl0124's profile pic

ayl0124 | Student, Grade 12 | (Level 1) Valedictorian

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First, you need to find the mass of each element in the compound. You already know that nitrogen makes up 30.5% of the compound.

`91.8 xx 0.305 = 27.999`

Oxygen makes up the remainder of the mass. 

`91.8 - 27.999 = 63.801`

Now, you can find the number of moles of nitrogen and oxygen in the compound.

One mole of nitrogen is 14 grams. 

`"27.999g" = ("1mol")/("14g") = "1.99mol"`

One mole of oxygen is 16 grams.

`"63.801g" = ("1mol")/("16g") = "3.98mol"`

The molecular formula of the compound that weighs 91.8 grams would be after rounding the moles to whole numbers:

`"N_2O_4"`

The empirical formula is the simplest ratio between elements. Both nitrogen's and oxygen's subscripts can be divided by a common factor of two. Therefore, the empirical formula is:

`"NO_2"` 

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