What is the empirical formula for a sample containing 81.70% carbon and 18.29% hydrogen?

Group of answer choices

CH4

C2H6

C3H8

CH

CH3

To determine the empirical formula, we follow these steps:

1. **Convert the percentages to grams**:
   - Assume we have 100 grams of the sample, so the mass of carbon is 81.70 grams and the mass of hydrogen is 18.29 grams.

2. **Convert grams to moles**:
   - Moles of carbon (C) = \( \frac{81.70 \text{ g}}{12.01 \text{ g/mol}} \approx 6.81 \text{ mol} \)
   - Moles of hydrogen (H) = \( \frac{18.29 \text{ g}}{1.008 \text{ g/mol}} \approx 18.14 \text{ mol} \)

3. **Find the simplest ratio**:
   - Divide both values by the smallest number of moles (6.81):
     - Carbon: \( \frac{6.81}{6.81} = 1 \)
     - Hydrogen: \( \frac{18.14}{6.81} \approx 2.66 \)

4. **Round to the nearest whole number**:
   - Since 2.66 is close to 3, we approximate it as 3.

Thus, the simplest ratio of C to H is 1:3, and the **empirical formula is CH₃**.

### Answer: **CH₃**

Would you like further details on any of the steps? Here are five related questions:

1. How do you determine molecular formulas from empirical formulas?
2. What is the difference between empirical and molecular formulas?
3. How does rounding affect empirical formula determination?
4. What role do molar masses play in finding empirical formulas?
5. How would the empirical formula change if oxygen were involved?

**Tip:** Always round carefully when dealing with ratios in empirical formula calculations!

What is the empirical formula for a sample containing 81.70% carbon and 18.29% hydrogen?

Find the empirical formula for a compound with 81.70% carbon and 18.29% hydrogen. This problem demonstrates how to convert mass percentages to moles, determine mole ratios, and simplify to find the empirical formula. Suitable for Grades 10-12 chemistry students.

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