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

Question

Accepted Answer

### Problem 1
# Explanation:
## Step1: Assume 100g of the compound.
Mass of C = 48.38g, Mass of H = 8.12g, Mass of O = $100 - 48.38 - 8.12 = 43.5$g.
## Step2: Convert mass to moles.
Moles of C: $\frac{48.38\mathrm{g}}{12.01\mathrm{g/mol}} \approx 4.028$mol
Moles of H: $\frac{8.12\mathrm{g}}{1.008\mathrm{g/mol}} \approx 8.056$mol
Moles of O: $\frac{43.5\mathrm{g}}{16.00\mathrm{g/mol}} \approx 2.719$mol
## Step3: Divide by the smallest (O's moles).
C: $\frac{4.028}{2.719} \approx 1.48$ (≈1.5), H: $\frac{8.056}{2.719} \approx 2.96$ (≈3), O: $\frac{2.719}{2.719} = 1$
## Step4: Multiply by 2 to get whole numbers.
C: $1.5\times2 = 3$, H: $3\times2 = 6$, O: $1\times2 = 2$
# Answer: $C_3H_6O_2$

### Problem 2
# Explanation:
## Step1: Assume 100g of the compound.
Mass of N = 46.67g, H = 6.70g, C = 19.98g, O = 26.65g.
## Step2: Convert to moles.
Moles of N: $\frac{46.67\mathrm{g}}{14.01\mathrm{g/mol}} \approx 3.331$mol
Moles of H: $\frac{6.70\mathrm{g}}{1.008\mathrm{g/mol}} \approx 6.647$mol
Moles of C: $\frac{19.98\mathrm{g}}{12.01\mathrm{g/mol}} \approx 1.663$mol
Moles of O: $\frac{26.65\mathrm{g}}{16.00\mathrm{g/mol}} \approx 1.666$mol
## Step3: Divide by the smallest (C or O, ≈1.66).
N: $\frac{3.331}{1.663} \approx 2$, H: $\frac{6.647}{1.663} \approx 4$, C: $\frac{1.663}{1.663} = 1$, O: $\frac{1.666}{1.663} \approx 1$
# Answer: $C_1H_4N_2O_1$ (or $CH_4N_2O$)

### Problem 3
# Explanation:
## Step1: Calculate empirical formula mass (CH).
Empirical mass = $12.01 + 1.008 = 13.018\mathrm{g/mol}$
## Step2: Find the ratio $n = \frac{\text{Molar Mass}}{\text{Empirical Mass}}$.
$n = \frac{78.11\mathrm{g/mol}}{13.018\mathrm{g/mol}} \approx 6$
## Step3: Multiply empirical formula by $n$.
Molecular formula: $ (CH)_6 = C_6H_6 $
# Answer: $C_6H_6$

### Problem 4
# Explanation:
## Step1: Empirical mass (CH) = $13.018\mathrm{g/mol}$
## Step2: Calculate $n = \frac{26.04\mathrm{g/mol}}{13.018\mathrm{g/mol}} \approx 2$
## Step3: Molecular formula: $ (CH)_2 = C_2H_2 $
# Answer: $C_2H_2$

### Problem 5
# Explanation:
## Step1: Convert masses to moles.
Moles of N: $\frac{1.121\mathrm{g}}{14.01\mathrm{g/mol}} \approx 0.0800$mol
Moles of H: $\frac{0.161\mathrm{g}}{1.008\mathrm{g/mol}} \approx 0.1597$mol
Moles of C: $\frac{0.480\mathrm{g}}{12.01\mathrm{g/mol}} \approx 0.03997$mol
Moles of O: $\frac{0.640\mathrm{g}}{16.00\mathrm{g/mol}} = 0.0400$mol
## Step2: Divide by the smallest (C or O, ≈0.04).
N: $\frac{0.0800}{0.03997} \approx 2$, H: $\frac{0.1597}{0.03997} \approx 4$, C: $\frac{0.03997}{0.03997} = 1$, O: $\frac{0.0400}{0.03997} \approx 1$
# Answer: $CH_4N_2O$ (same as Problem 2, due to rounding)

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QUESTION IMAGE

Question

Explanation:

Problem 1

Step1: Assume 100g of the compound.

Step2: Convert mass to moles.

Step3: Divide by the smallest (O's moles).

Step4: Multiply by 2 to get whole numbers.

Step1: Assume 100g of the compound.

Step2: Convert to moles.

Step3: Divide by the smallest (C or O, ≈1.66).

Step1: Calculate empirical formula mass (CH).

Step2: Find the ratio \(n = \frac{\text{Molar Mass}}{\text{Empirical Mass}}\).

Step3: Multiply empirical formula by \(n\).

Answer:

Problem 2