Answered

Select the correct answer from each dropdown menu.

Determine the molecular formula of the compound.

A compound is used as a gasoline additive. It has a molecular weight of 60.10 atomic mass units and an empirical formula of [tex]$C_3H_6O$[/tex]. The molecular formula is [tex]$C_3H_6O$[/tex].



Answer :

GinnyAnswer
Let's work through this problem step-by-step to determine the molecular formula based on the given information:

1. Empirical Formula Mass Calculation:
- The empirical formula given is [tex]\( C_3H_1O_1 \)[/tex].
- The atomic masses are approximately:
- Carbon (C): 12.01 atomic mass units (amu)
- Hydrogen (H): 1.01 amu
- Oxygen (O): 16.00 amu
- Calculate the empirical formula mass:
[tex]\[ 3 \times 12.01 \, \text{(C)} + 1 \times 1.01 \, \text{(H)} + 1 \times 16.00 \, \text{(O)} = 36.03 + 1.01 + 16.00 = 53.04 \, \text{amu} \][/tex]

2. Determine the Multiplier for the Molecular Formula:
- The molecular weight provided is 60.10 amu.
- To find the multiplier (n), divide the molecular weight by the empirical formula mass:
[tex]\[ n = \frac{60.10 \, \text{amu}}{53.04 \, \text{amu}} \approx 1 \][/tex]
- Since the result is approximately 1, the molecular formula is the same as the empirical formula.

3. Calculate the Total Number of Atoms in the Molecular Formula:
- For carbon (C): [tex]\( 3 \times 1 = 3 \)[/tex]
- For hydrogen (H): [tex]\( 1 \times 1 = 1 \)[/tex]
- For oxygen (O): [tex]\( 1 \times 1 = 1 \)[/tex]

So, the molecular formula for the compound is [tex]\( C_3H_1O_1 \)[/tex]. This can be formatted as [tex]\( C_3H_1O \)[/tex].

Hence, the molecular formula is:
[tex]\[ C_3H_1O \][/tex]

Other Questions