nails4life324

03.06.2021 •

Chemistry

When a very massive star runs out of elements to fuse, it

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shawna189

20.10.2022

Chemistry

Empirical formula is NO_2 and molecular formula is N_2O_4

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

Step 1: 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 = $\frac{69.54}{100}\times 100g=69.54$

Step 2: 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 = $\frac{69.54g}{16g/mol}=4.35mol$

Step 3: 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 = $\frac{4.35}{2.18}=1.99\approx 2$

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

Empirical formula = N_1O_2=NO_2

Step 5: 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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