8.84 g of a compound is analyzed and found to contain 3.53 g of C, 0.295 g of H, 4.08 g of N, and 0.935 g of O. Which of the following could be the molecular formula for this compound?

C5H10N3O3

1. C10H10N10O2

2. C2H4N6O

3. C8H8N8O4

4. C4H8N6O2

Option 1 is correct. C10H10N10O2

Explanation:

Step 1: Data given

Mass of a compound = 8.84 grams

Mass of C = 3.53 grams

Mass of H = 0.295 grams

Mass of N = 4.08 grams

Mass of O = 0.935 grams

Atomic mass C = 12.01 g/mol

Atomic mass H = 1.01 g/mol

Atomic mass N = 14.0 g/mol

Atomic mass O = 16.0 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = 3.53 grams / 12.01 g/mol

Moles C = 0.294 moles

Moles H = 0.295 grams / 1.01 g/mol

Moles H = 0.292 moles

Moles N = 4.08 grams / 14.0 g/mol

Moles N = 0.291 moles

Moles O = 0.935 grams / 16 g/mol

Moles O = 0.0584 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.294 / 0.0584 = 5

H: 0.292 / 0.0584 = 5

N: 0.291/0.0584 = 5

O: 0.0584/0.0584 = 1

The empirical formula is C5H5N5O

Option 1 has this empirical formula

C10H10N10O2