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. Then mass of C = 48.38g, H = 8.12g, O = 100 - 48.38 - 8.12 = 43.5g.
## Step2: Convert mass to moles. Moles of C: $\frac{48.38}{12.01} \approx 4.03$; Moles of H: $\frac{8.12}{1.008} \approx 8.06$; Moles of O: $\frac{43.5}{16.00} \approx 2.72$.
## Step3: Divide by the smallest (2.72). C: $\frac{4.03}{2.72} \approx 1.48 \approx 1.5$; H: $\frac{8.06}{2.72} \approx 2.96 \approx 3$; O: $\frac{2.72}{2.72} = 1$.
## Step4: Multiply by 2 to get whole numbers. C: 3, H: 6, O: 2.
# Answer: $\ce{C3H6O2}$

### Problem 2
# Explanation:
## Step1: Assume 100g. Masses: N = 46.67g, H = 6.70g, C = 19.98g, O = 26.65g.
## Step2: Moles. N: $\frac{46.67}{14.01} \approx 3.33$; H: $\frac{6.70}{1.008} \approx 6.65$; C: $\frac{19.98}{12.01} \approx 1.66$; O: $\frac{26.65}{16.00} \approx 1.66$.
## Step3: Divide by smallest (1.66). N: $\frac{3.33}{1.66} \approx 2$; H: $\frac{6.65}{1.66} \approx 4$; C: $\frac{1.66}{1.66} = 1$; O: $\frac{1.66}{1.66} = 1$.
# Answer: $\ce{C1H4N2O1}$ (or $\ce{CH4N2O}$)

### Problem 3
# Explanation:
## Step1: Empirical formula mass of CH: 12.01 + 1.008 = 13.018 g/mol.
## Step2: Find the ratio $n = \frac{	ext{Molar Mass}}{	ext{Empirical Mass}} = \frac{78.11}{13.018} \approx 6$.
## Step3: Multiply empirical formula by n. CH * 6 = $\ce{C6H6}$.
# Answer: $\ce{C6H6}$

### Problem 4
# Explanation:
## Step1: Empirical formula mass of CH: 13.018 g/mol.
## Step2: Ratio $n = \frac{26.04}{13.018} \approx 2$.
## Step3: Multiply CH by 2: $\ce{C2H2}$.
# Answer: $\ce{C2H2}$

### Problem 5
# Explanation:
## Step1: Convert masses to moles. N: $\frac{1.121}{14.01} \approx 0.08$; H: $\frac{0.161}{1.008} \approx 0.16$; C: $\frac{0.480}{12.01} \approx 0.04$; O: $\frac{0.640}{16.00} \approx 0.04$.
## Step2: Divide by smallest (0.04). N: $\frac{0.08}{0.04} = 2$; H: $\frac{0.16}{0.04} = 4$; C: $\frac{0.04}{0.04} = 1$; O: $\frac{0.04}{0.04} = 1$.
# Answer: $\ce{C1H4N2O1}$ (or $\ce{CH4N2O}$)

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QUESTION IMAGE

Question

Explanation:

Problem 1

Step1: Assume 100g of the compound. Then mass of C = 48.38g, H = 8.12g, O = 100 - 48.38 - 8.12 = 43.5g.

Step2: Convert mass to moles. Moles of C: $\frac{48.38}{12.01} \approx 4.03$; Moles of H: $\frac{8.12}{1.008} \approx 8.06$; Moles of O: $\frac{43.5}{16.00} \approx 2.72$.

Step3: Divide by the smallest (2.72). C: $\frac{4.03}{2.72} \approx 1.48 \approx 1.5$; H: $\frac{8.06}{2.72} \approx 2.96 \approx 3$; O: $\frac{2.72}{2.72} = 1$.

Step4: Multiply by 2 to get whole numbers. C: 3, H: 6, O: 2.

Step1: Assume 100g. Masses: N = 46.67g, H = 6.70g, C = 19.98g, O = 26.65g.

Step2: Moles. N: $\frac{46.67}{14.01} \approx 3.33$; H: $\frac{6.70}{1.008} \approx 6.65$; C: $\frac{19.98}{12.01} \approx 1.66$; O: $\frac{26.65}{16.00} \approx 1.66$.

Step3: Divide by smallest (1.66). N: $\frac{3.33}{1.66} \approx 2$; H: $\frac{6.65}{1.66} \approx 4$; C: $\frac{1.66}{1.66} = 1$; O: $\frac{1.66}{1.66} = 1$.

Step1: Empirical formula mass of CH: 12.01 + 1.008 = 13.018 g/mol.

Step2: Find the ratio $n = \frac{\text{Molar Mass}}{\text{Empirical Mass}} = \frac{78.11}{13.018} \approx 6$.

Step3: Multiply empirical formula by n. CH * 6 = $\ce{C6H6}$.

Answer:

Problem 2