Question

By elemental analysis of a biomass sample, it was found that the cells in the given sample are composed of (by weight) C=50%, H=8%, O=20%, N=10% and ash content=12%. Which of the following is the correct empirical formula of the ash-free cells, normalized with respect to Nitrogen?

• A. C_5.8H_11.2O_1.7N
• B. C_5.0H_10.2O_1.1N
• C. C_6.8H_12.2O_2.1N
• D. C_4.8H_9.2O_3.1N

Hint:

Convert each weight percent to moles using atomic weights, then divide every mole value by the moles of nitrogen.

Answer 1

The Correct Option is A

Step 1: Set up a convenient basis.
Take 100 g of the biomass sample as the basis. By weight this gives: carbon = 50 g, hydrogen = 8 g, oxygen = 20 g, nitrogen = 10 g, and ash = 12 g. Note that \(50+8+20+10+12=100\), so these five numbers already account for the whole sample; the ash-free (organic) part is just the \(C, H, O, N\) part.

Step 2: Convert each element's mass to moles.
Use atomic weights \(C=12.0\), \(H=1.0\), \(O=16.0\), \(N=14.0\) g/mol:
\[ n_C = \frac{50}{12.0} = 4.17 \text{ mol}, \quad n_H = \frac{8}{1.0} = 8.0 \text{ mol} \]
\[ n_O = \frac{20}{16.0} = 1.25 \text{ mol}, \quad n_N = \frac{10}{14.0} = 0.714 \text{ mol} \]

Step 3: Normalize every mole value to nitrogen.
The empirical formula must be written "normalized with respect to Nitrogen," so we divide every mole count by \(n_N\), making nitrogen's own coefficient equal to 1:
\[ \frac{n_C}{n_N} = \frac{4.17}{0.714} = 5.83, \quad \frac{n_H}{n_N} = \frac{8.0}{0.714} = 11.2 \]
\[ \frac{n_O}{n_N} = \frac{1.25}{0.714} = 1.75, \quad \frac{n_N}{n_N} = 1 \]

Step 4: Write the empirical formula and match with the options.
The ash-free, nitrogen-normalized empirical formula is:
\[ C_{5.8}H_{11.2}O_{1.7}N \]
This matches option (A) exactly. The other options use different atomic ratios: option (B) has too low an oxygen ratio, option (C) has both carbon and oxygen too high, and option (D) has carbon too low and oxygen too high, none of which come from dividing the actual mole numbers above by \(n_N\).

Final Answer:
The ash-free empirical formula normalized to nitrogen is \(C_{5.8}H_{11.2}O_{1.7}N\), which is option (A).
\[ \boxed{C_{5.8}H_{11.2}O_{1.7}N} \]