Question

empirical formula & molecular formula ws
remember: percent to mass, mass to moles, divide by the smallest, & (if necessary) multiply til whole

1. a compound is found to have (by mass) 48.38% carbon, 8.12% hydrogen and the rest oxygen. what is its empirical formula?
2. a compound is found to have 46.67% nitrogen, 6.70% hydrogen, 19.98% carbon and 26.65% oxygen. what is its empirical formula?
3. a compound is known to have an empirical formula of ch and a molar mass of 78.11 g/mol. what is its molecular formula?
4. another compound, also with an empirical formula if ch is found to have a molar mass of 26.04 g/mol. what is its molecular formula?
5. a compound is found to have 1.121 g nitrogen, 0.161 g hydrogen, 0.480 g carbon and 0.640 g oxygen. what is its empirical formula? (note that masses are given, not percentages.)

Explanation:

Let's solve question 1 as an example (we can follow the same steps for other questions too).

Question 1:

A compound is found to have (by mass) 48.38% carbon, 8.12% hydrogen and the rest oxygen. What is its empirical formula?

Step 1: Assume 100g of the compound

If we assume we have 100g of the compound, then the masses of each element are:

• Carbon (C): \( 48.38\ \text{g} \)
• Hydrogen (H): \( 8.12\ \text{g} \)
• Oxygen (O): \( 100 - 48.38 - 8.12 = 43.5\ \text{g} \)

Step 2: Convert masses to moles

Molar mass of C: \( 12.01\ \text{g/mol} \), H: \( 1.008\ \text{g/mol} \), O: \( 16.00\ \text{g/mol} \)

• Moles of C: \( \frac{48.38\ \text{g}}{12.01\ \text{g/mol}} \approx 4.028\ \text{mol} \)
• Moles of H: \( \frac{8.12\ \text{g}}{1.008\ \text{g/mol}} \approx 8.056\ \text{mol} \)
• Moles of O: \( \frac{43.5\ \text{g}}{16.00\ \text{g/mol}} \approx 2.719\ \text{mol} \)

Step 3: Divide by the smallest number of moles

The smallest number of moles is approximately \( 2.719\ \text{mol} \) (from O).

• Ratio of C: \( \frac{4.028}{2.719} \approx 1.48 \approx 1.5 \) (or \( \frac{3}{2} \))
• Ratio of H: \( \frac{8.056}{2.719} \approx 2.96 \approx 3 \)
• Ratio of O: \( \frac{2.719}{2.719} = 1 \)

Step 4: Multiply by a factor to get whole numbers

Since we have a ratio of \( 1.5:3:1 \) (or \( \frac{3}{2}:3:1 \)), multiply all by 2 to eliminate the fraction:

• C: \( 1.5 \times 2 = 3 \)
• H: \( 3 \times 2 = 6 \)
• O: \( 1 \times 2 = 2 \)

Step 1: Assume 100g of the compound

Masses:

• N: \( 46.67\ \text{g} \)
• H: \( 6.70\ \text{g} \)
• C: \( 19.98\ \text{g} \)
• O: \( 26.65\ \text{g} \)

Step 2: Convert to moles

Molar masses: N=14.01, H=1.008, C=12.01, O=16.00

• Moles of N: \( \frac{46.67}{14.01} \approx 3.331\ \text{mol} \)
• Moles of H: \( \frac{6.70}{1.008} \approx 6.647\ \text{mol} \)
• Moles of C: \( \frac{19.98}{12.01} \approx 1.663\ \text{mol} \)
• Moles of O: \( \frac{26.65}{16.00} \approx 1.666\ \text{mol} \)

Step 3: Divide by the smallest number of moles

The smallest number of moles is approximately \( 1.663\ \text{mol} \) (from C, close to O's 1.666). Let's use 1.663 (or 1.666, they are almost the same).

• Ratio of N: \( \frac{3.331}{1.663} \approx 2.0 \)
• Ratio of H: \( \frac{6.647}{1.663} \approx 4.0 \)
• Ratio of C: \( \frac{1.663}{1.663} = 1.0 \)
• Ratio of O: \( \frac{1.666}{1.663} \approx 1.0 \)

Step 1: Calculate the molar mass of the empirical formula

Empirical formula: \( CH \)
Molar mass of \( CH \): \( 12.01 + 1.008 = 13.018\ \text{g/mol} \)

Step 2: Determine the ratio (n) of molecular mass to empirical mass

\( n = \frac{\text{Molar mass of compound}}{\text{Molar mass of empirical formula}} = \frac{78.11\ \text{g/mol}}{13.018\ \text{g/mol}} \approx 6 \)

Step 3: Multiply the empirical formula by n

Molecular formula = \( (CH)_n = (CH)_6 = C_6H_6 \)

Answer:

The empirical formula is \( \boldsymbol{C_3H_6O_2} \) (Wait, wait, let's check the calculations again. Wait, maybe I made a mistake in the oxygen mass. Wait, 100 - 48.38 - 8.12 = 43.5? Wait, 48.38 + 8.12 = 56.5, so 100 - 56.5 = 43.5. That's correct. Then moles of O: 43.5 / 16 ≈ 2.71875. Moles of C: 48.38 / 12.01 ≈ 4.028. 4.028 / 2.71875 ≈ 1.48. 8.12 / 1.008 ≈ 8.055. 8.055 / 2.71875 ≈ 2.96. Hmm, maybe a better approach: let's use more precise calculations.

Wait, 48.38 g C: 48.38 / 12.01 = 4.028 mol C.
8.12 g H: 8.12 / 1.008 = 8.055 mol H.
43.5 g O: 43.5 / 16 = 2.71875 mol O.

Now, divide each by 2.71875:

C: 4.028 / 2.71875 ≈ 1.481 (≈ 1.5)
H: 8.055 / 2.71875 ≈ 2.963 (≈ 3)
O: 1

So, to get whole numbers, multiply by 2:

C: 1.5 * 2 = 3
H: 3 * 2 = 6
O: 1 * 2 = 2

So the empirical formula is \( C_3H_6O_2 \)? Wait, but let's check the percentages. Let's calculate the mass percentages of \( C_3H_6O_2 \):

Molar mass of \( C_3H_6O_2 \): 312.01 + 61.008 + 2*16.00 = 36.03 + 6.048 + 32.00 = 74.078 g/mol

% C: (36.03 / 74.078) * 100 ≈ 48.64% (close to 48.38, considering rounding errors)
% H: (6.048 / 74.078) * 100 ≈ 8.16% (close to 8.12)
% O: (32.00 / 74.078) * 100 ≈ 43.20% (close to 43.5, again rounding errors). So that's acceptable.

Alternatively, maybe the empirical formula is \( C_2H_4O \)? Wait, no, let's try again. Wait, maybe I messed up the oxygen calculation. Wait, 48.38% C, 8.12% H, so O is 43.5%. Let's use moles:

C: 48.38 / 12.01 = 4.028 mol
H: 8.12 / 1.008 = 8.055 mol
O: 43.5 / 16 = 2.71875 mol

Now, divide by the smallest (2.71875):

C: 4.028 / 2.71875 ≈ 1.48 (≈ 3/2)
H: 8.055 / 2.71875 ≈ 2.96 (≈ 3)
O: 1

So multiplying by 2: C=3, H=6, O=2. So \( C_3H_6O_2 \) is correct.

Question 2:

A compound is found to have 46.67% nitrogen, 6.70% hydrogen, 19.98% carbon and 26.65% oxygen. What is its empirical formula?