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Vsevolod [243]
3 years ago
8

A compound of nitrogen and oxygen is 30.46 percent by mass of N and 69.54 percent by mass O. The molar mass if the compound was

determined to be 92g/mol. what is the emperical formula and molecular formular
Chemistry
1 answer:
agasfer [191]3 years ago
6 0

Answer: Empirical formula is NO_2 and molecular formula is N_2O_4

Explanation: To find the empirical and molecular formula, we follow few steps:

<u>Step 1: </u>Converting mass percent into mass

We are given the percentage of elements by mass. So, the total mass take will be 100 grams.

Therefore, mass of nitrogen = \frac{30.46}{100}\times 100g=30.46g

Similarly, mass of oxygen = \frac{69.54}{100}\times 100g=69.54

<u>Step 2:</u> Converting the masses into their respective moles

We use the formula:

Moles=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of Nitrogen = 14 g/mol

Molar mass of oxygen = 16 g/mol

Moles of nitrogen = \frac{30.46g}{14g/mol}=2.18moles

Moles of oxygen = \frac{69.54g}{16g/mol}=4.35mol

<u>Step 3: </u>Getting the mole ratio of nitrogen and oxygen by dividing the calculated moles by the lowest mole value.

Mole ratio of nitrogen = \frac{2.18}{2.18}=1

Mole ratio of oxyegn = \frac{4.35}{2.18}=1.99\approx 2

<u>Step 4: </u>The mole ratio of elements are represented as the subscripts in a empirical formula, we get

Empirical formula = N_1O_2=NO_2

<u>Step 5: </u>For molecular formula, we divide the molar mass of the compound by the empirical molar mass.

Empirical molar mass of NO_2=(14\times 1)+(16\times 2)g/mol

Empirical molar mass = 46 g/mol

Molar mass of the compound = 92 g/mol

n=\frac{\text{Molar mass}}{\text{Empirical molar mass}}

n=\frac{92g/mol}{46g/mol}=2

Now, multiplying each of the subscript of empirical formula by 'n', we get

Molecular formula = N_2O_4

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