A substance contains 5.00 g C and 1.70 g H. What is the empirical formula? (Answer starts with C and H is last.)

To determine the empirical formula of the substance, we will follow these steps:

1. **Find the moles of Carbon (C):**
   The molar mass of carbon (C) is approximately 12.01 g/mol.

   \[
   \text{Moles of C} = \frac{5.00 \, \text{g}}{12.01 \, \text{g/mol}} = 0.416 \, \text{mol}
   \]

2. **Find the moles of Hydrogen (H):**
   The molar mass of hydrogen (H) is approximately 1.008 g/mol.

   \[
   \text{Moles of H} = \frac{1.70 \, \text{g}}{1.008 \, \text{g/mol}} = 1.686 \, \text{mol}
   \]

3. **Determine the simplest mole ratio:**
   Now, divide both mole quantities by the smallest number of moles (which is 0.416 mol).

   For carbon:
   \[
   \frac{0.416}{0.416} = 1
   \]

   For hydrogen:
   \[
   \frac{1.686}{0.416} \approx 4.05 \quad (\text{approximately } 4)
   \]

4. **Write the empirical formula:**
   The simplest whole number ratio of carbon to hydrogen is 1:4, so the empirical formula is:

   \[
   \mathbf{CH_4}
   \]

The empirical formula of the substance is **CH₄**.

Would you like further details on any part of the process?

Here are 5 related questions:
1. How do you calculate molecular formulas from empirical formulas?
2. What is the difference between an empirical formula and a molecular formula?
3. How do you find empirical formulas for substances with more than two elements?
4. How do you account for rounding errors in empirical formula calculations?
5. How does the empirical formula help in determining chemical structure?

**Tip:** Always verify that the mole ratio simplifies to the nearest whole numbers for an accurate empirical formula.

Learn how to calculate the empirical formula of a substance containing 5.00 g of carbon and 1.70 g of hydrogen. This example demonstrates key stoichiometric concepts, including mole calculations and the empirical formula determination. Suitable for students in Grades 9-12.

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