Acetaldehyde has the molecular formula C2H4O. Given 55.0 g of Acetaldehyde answer the following questions.
Find the number of moles of hydrogen in 55.0 g of acetaldehyde (C2H4O) and molecules?
 Find the number of carbon atoms in of 55.0 g of acetaldehyde (C2H4O)?
The Correct Answer and Explanation is :
For 55.0 g of acetaldehyde (C2H4O), here are the calculations:
- Moles of acetaldehyde:
The molecular weight of acetaldehyde (C2H4O) is 44.05 g/mol. Given 55.0 g of acetaldehyde, we calculate the number of moles of acetaldehyde:
[
\text{Moles of acetaldehyde} = \frac{55.0 \, \text{g}}{44.05 \, \text{g/mol}} \approx 1.25 \, \text{moles}
] - Moles of hydrogen:
Each molecule of acetaldehyde contains 4 hydrogen atoms. So, the number of moles of hydrogen in 1.25 moles of acetaldehyde is:
[
\text{Moles of hydrogen} = 4 \times 1.25 = 5.00 \, \text{moles}
] - Molecules of hydrogen:
Using Avogadro’s number (6.022 × 10²³ molecules per mole), the number of hydrogen molecules is:
[
\text{Molecules of hydrogen} = 5.00 \, \text{moles} \times 6.022 \times 10^{23} \, \text{molecules/mole} \approx 3.01 \times 10^{24} \, \text{molecules}
] - Moles of carbon:
Each molecule of acetaldehyde contains 2 carbon atoms. Therefore, the number of moles of carbon in 1.25 moles of acetaldehyde is:
[
\text{Moles of carbon} = 2 \times 1.25 = 2.50 \, \text{moles}
] - Atoms of carbon:
Using Avogadro’s number, the number of carbon atoms in 2.50 moles of acetaldehyde is:
[
\text{Atoms of carbon} = 2.50 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mole} \approx 1.50 \times 10^{24} \, \text{atoms}
]
To summarize:
- Moles of hydrogen: 5.00 moles
- Molecules of hydrogen: 3.01 × 10²⁴ molecules
- Moles of carbon: 2.50 moles
- Atoms of carbon: 1.50 × 10²⁴ atoms
Let me generate an image to visualize this calculation.
Here is the scientific diagram that visualizes the molecular composition of acetaldehyde (C2H4O), including the number of moles of hydrogen, carbon, and the corresponding number of molecules and atoms. This should help in understanding the calculations! Let me know if you need further explanations or modifications.
