Understanding Stoichiometry and Moles
To solve this problem, we need to bridge the gap between mass and the number of atoms using the concept of the mole. A mole represents $6.022 \times 10^{23}$ particles (Avogadro's number).
Step 1: Find the number of moles of Methane ($CH_4$)
First, we determine the molar mass of methane ($CH_4$):
- Carbon ($C$) = 12.01 g/mol
- Hydrogen ($H$) = 1.008 g/mol
- Molar Mass of $CH_4 = 12.01 + 4(1.008) \approx 16.04$ g/mol (often rounded to 16 g/mol for calculation).
Now, calculate the moles of $CH_4$ in 32 g: $$\text{Moles of } CH_4 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{32 \text{ g}}{16 \text{ g/mol}} = 2 \text{ moles of } CH_4$$
Step 2: Determine the number of Hydrogen atoms
Each molecule of methane ($CH_4$) contains 4 hydrogen atoms. Therefore, 2 moles of $CH_4$ will contain: $$\text{Moles of } H \text{ atoms} = 2 \text{ moles of } CH_4 \times 4 = 8 \text{ moles of } H \text{ atoms}$$
Step 3: Relate to Water ($H_2O$)
We need to find the mass of water ($H_2O$) that also contains 8 moles of hydrogen atoms.
Each molecule of water ($H_2O$) contains 2 hydrogen atoms. Let $n$ be the number of moles of water molecules needed: $$\text{Moles of } H \text{ atoms} = n \times 2$$ $$8 = n \times 2 \implies n = 4 \text{ moles of } H_2O$$
Step 4: Calculate the mass of Water
Finally, we convert the moles of water back into mass. The molar mass of water ($H_2O$) is:
- $2(1.008) + 16.00 \approx 18.02$ g/mol (often rounded to 18 g/mol).
$$\text{Mass of } H_2O = \text{moles} \times \text{molar mass}$$ $$\text{Mass of } H_2O = 4 \text{ moles} \times 18 \text{ g/mol} = 72 \text{ g}$$
Summary
By calculating the total amount of hydrogen present in 32g of methane, we determined we needed 8 moles of hydrogen. Since each water molecule contains 2 hydrogen atoms, we required 4 moles of water, which equates to 72 grams.