Question

1. what is the empirical formula of a molecule containing 14.7% carbon, 1.9% hydrogen, 49.0% oxygen, and 34.4% nitrogen by mass?

symbol | percentage | percent as mass | divide by small | round or multiply | subscripts

answer =

Explanation:

Step1: Assume 100g sample, set masses

Let total mass = 100g. So:
$m_C = 14.7\ \text{g}$, $m_H = 1.9\ \text{g}$, $m_O = 49.0\ \text{g}$, $m_N = 34.4\ \text{g}$

Step2: Calculate moles of each element

Moles $n = \frac{\text{mass}}{\text{molar mass}}$, molar masses: $M_C=12.01\ \text{g/mol}$, $M_H=1.008\ \text{g/mol}$, $M_O=16.00\ \text{g/mol}$, $M_N=14.01\ \text{g/mol}$
$n_C = \frac{14.7}{12.01} \approx 1.224\ \text{mol}$
$n_H = \frac{1.9}{1.008} \approx 1.885\ \text{mol}$
$n_O = \frac{49.0}{16.00} = 3.0625\ \text{mol}$
$n_N = \frac{34.4}{14.01} \approx 2.455\ \text{mol}$

Step3: Divide by smallest mole value

Smallest $n = 1.224\ \text{mol}$
$C: \frac{1.224}{1.224} = 1$
$H: \frac{1.885}{1.224} \approx 1.54$
$O: \frac{3.0625}{1.224} \approx 2.50$
$N: \frac{2.455}{1.224} \approx 2.006$

Step4: Multiply to get whole numbers

Multiply all by 2 to eliminate decimals:
$C: 1 \times 2 = 2$
$H: 1.54 \times 2 \approx 3$
$O: 2.50 \times 2 = 5$
$N: 2.006 \times 2 \approx 4$

Completed Table:

Symbol	Percentage	Percent As Mass	Divide by Small	Round or Multiply	Subscripts
H	1.9%	1.9 g	≈1.54	×2	3
O	49.0%	49.0 g	≈2.50	×2	5
N	34.4%	34.4 g	≈2.006	×2	4

Answer:

$\text{C}_2\text{H}_3\text{O}_5\text{N}_4$