QuestionJuly 5, 2025

B) A Compound has the following percent cosition by mass: 42.87% carbon, 3.598% hydrogen, 28.55% oxygen, and 25.00% nitrogen. If the molar mass of the compound is known to be between 165-170g/mol determine the empirical and molecular Moles of C: square moles C Moles of H: square moles H Moles of 0: square moles 0 Moles of N: square moles N Empinical Formula: square Molecular Formula: square

B) A Compound has the following percent cosition by mass: 42.87% carbon, 3.598% hydrogen, 28.55% oxygen, and 25.00% nitrogen. If the molar mass of the compound is known to be between 165-170g/mol determine the empirical and molecular Moles of C: square moles C Moles of H: square moles H Moles of 0: square moles 0 Moles of N: square moles N Empinical Formula: square Molecular Formula: square
B)
A Compound has the following percent cosition by mass:
42.87%  carbon, 3.598%  hydrogen, 28.55%  oxygen, and 25.00% 
nitrogen. If the molar mass of the compound is known to be between
165-170g/mol
determine the empirical and molecular
Moles of C: square  moles C
Moles of H: square  moles H
Moles of 0: square  moles 0
Moles of N: square  moles N
Empinical Formula: square 
Molecular Formula: square

Solution
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Answer

Moles of C: 3.57 moles C ### Moles of H: 3.57 moles H ### Moles of O: 1.78 moles O ### Moles of N: 1.78 moles N ### Empirical Formula: C_2H_2ON ### Molecular Formula: C_6H_6O_3N_3 Explanation 1. Calculate moles of each element Assume 100g of compound. Moles of C = \frac{42.87}{12.01} \approx 3.57, H = \frac{3.598}{1.008} \approx 3.57, O = \frac{28.55}{16.00} \approx 1.78, N = \frac{25.00}{14.01} \approx 1.78. 2. Determine simplest mole ratio Divide by smallest number of moles (1.78): C = \frac{3.57}{1.78} \approx 2, H = \frac{3.57}{1.78} \approx 2, O = \frac{1.78}{1.78} = 1, N = \frac{1.78}{1.78} = 1. Empirical formula is C_2H_2ON. 3. Calculate empirical formula mass Mass of C_2H_2ON = 2(12.01) + 2(1.008) + 16.00 + 14.01 = 56.06 g/mol. 4. Determine molecular formula Molecular formula mass is between 165-170 g/mol. Ratio = \frac{168}{56.06} \approx 3. Molecular formula is (C_2H_2ON)_3 = C_6H_6O_3N_3.

Explanation

1. Calculate moles of each element<br /> Assume 100g of compound. Moles of C = $\frac{42.87}{12.01} \approx 3.57$, H = $\frac{3.598}{1.008} \approx 3.57$, O = $\frac{28.55}{16.00} \approx 1.78$, N = $\frac{25.00}{14.01} \approx 1.78$.<br /><br />2. Determine simplest mole ratio<br /> Divide by smallest number of moles (1.78): C = $\frac{3.57}{1.78} \approx 2$, H = $\frac{3.57}{1.78} \approx 2$, O = $\frac{1.78}{1.78} = 1$, N = $\frac{1.78}{1.78} = 1$. Empirical formula is $C_2H_2ON$.<br /><br />3. Calculate empirical formula mass<br /> Mass of $C_2H_2ON = 2(12.01) + 2(1.008) + 16.00 + 14.01 = 56.06$ g/mol.<br /><br />4. Determine molecular formula<br /> Molecular formula mass is between 165-170 g/mol. Ratio = $\frac{168}{56.06} \approx 3$. Molecular formula is $(C_2H_2ON)_3 = C_6H_6O_3N_3$.
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