Question

the percentage composition of acetaldehyde is 54.5% C,9.2% H,and 36.3% O,and its molecular mass is 44 amu. Obtain the molecular formula of acetaldehyde

Answer 1

Answer on Question #43515 - Chemistry - Organic Chemistry

Question:

The percentage composition of acetaldehyde is 54.5% C, 9.2% H, and 36.3% O, and its molecular mass is 44 amu. Obtain the molecular formula of acetaldehyde.

Answer:

Change each percentage to an expression of the mass of each element in grams. That is 54.5% C becomes 54.5 g C, 9.2% H becomes 9.2 g H, and 36.3% O becomes 36.3 g O.

Convert the amount of each element in grams to its amount in moles

$$\left(\frac{54.5 \text{ g C}}{1}\right) \left(\frac{1 \text{ mol}}{12.01 \text{ g C}}\right) = 4.54 \text{ mol}$$ $$\left(\frac{9.2 \text{ g H}}{1}\right) \left(\frac{1 \text{ mol}}{1.008 \text{ g H}}\right) = 9.13 \text{ mol}$$ $$\left(\frac{36.3 \text{ g O}}{1}\right) \left(\frac{1 \text{ mol}}{16.00 \text{ g O}}\right) = 2.27 \text{ mol}$$

Divide each of the found values by the smallest of these values (2.27)

$$\frac{4.54 \text{ mol}}{2.27 \text{ mol}} = 2.00$$ $$\frac{9.13 \text{ mol}}{2.27 \text{ mol}} = 4.02$$ $$\frac{2.27 \text{ mol}}{2.27 \text{ mol}} = 1.00$$

Thus, the empirical formula of acetaldehyde is $C_2H_4O$. The molecular mass of acetaldehyde ($C_2H_4O$) is :

$$M(C_2H_4O) = 2*M(C) + 4*M(H) + M(O) = 2*12.01 + 4*1.008 + 16.00 = 44.05$$

Answer: $C_2H_4O$

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