Question

empty vial mass: \underline{6.1772} \
mass vial + water: \underline{6.2414} \
b. calculate the number of hydrogen atoms present in the drop of water.

Explanation:

To calculate the number of hydrogen atoms in a drop of water, we follow these steps:

Step 1: Determine the mass of water

First, find the mass of water by subtracting the empty vial mass from the mass of the vial + water. Let’s denote:

• Empty vial mass: \( m_{\text{vial}} = 6.1772 \, \text{g} \)
• Mass of vial + water: \( m_{\text{vial+water}} = 6.2414 \, \text{g} \)

Mass of water:

$$ m_{\text{water}} = m_{\text{vial+water}} - m_{\text{vial}} = 6.2414 - 6.1772 = 0.0642 \, \text{g} $$

Step 2: Moles of water (\( \text{H}_2\text{O} \))

The molar mass of \( \text{H}_2\text{O} \) is \( 18.015 \, \text{g/mol} \) (since \( \text{H} = 1.008 \, \text{g/mol} \), \( \text{O} = 16.00 \, \text{g/mol} \); \( 2(1.008) + 16.00 = 18.016 \, \text{g/mol} \), approximated to \( 18.015 \, \text{g/mol} \)).

Moles of \( \text{H}_2\text{O} \):

$$ n_{\text{H}_2\text{O}} = \frac{m_{\text{water}}}{\text{Molar mass of } \text{H}_2\text{O}} = \frac{0.0642 \, \text{g}}{18.015 \, \text{g/mol}} \approx 0.003563 \, \text{mol} $$

Step 3: Moles of hydrogen atoms

In 1 mole of \( \text{H}_2\text{O} \), there are 2 moles of \( \text{H} \) atoms. Thus:

$$ n_{\text{H}} = 2 \times n_{\text{H}_2\text{O}} = 2 \times 0.003563 \, \text{mol} \approx 0.007126 \, \text{mol} $$

Step 4: Number of hydrogen atoms (Avogadro’s number)

Avogadro’s number (\( N_A \)) is \( 6.022 \times 10^{23} \, \text{atoms/mol} \).

Number of \( \text{H} \) atoms:

$$ N_{\text{H}} = n_{\text{H}} \times N_A = 0.007126 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 4.29 \times 10^{21} \, \text{atoms} $$

Answer:

The number of hydrogen atoms in the drop of water is approximately \( \boldsymbol{4.3 \times 10^{21}} \) (or \( 4.29 \times 10^{21} \)) atoms.