Question

determine the empirical formula for the compound whose percentage composition of \h.\ input the subscript values in c o order according to the format.
format: input the subscripts in order as a single number starting from left to right for each element, including a 1 value as appropriate. for example, for a formula of c₂h₃o₂br₁ or c₂h₃o₂br, the input is 2321
a 40.68% c, 5.12% h, 54.20% o b 15.8% c, 84.2% s c 43% c, 57% o d 12.1% c, 16.2% o, 71.7% cl
e 50.04% c, 5.59% h, 44.37% o f 40.0% c, 6.7% h, 53.3% o g 45.8% s, 54.2% f h 30.45% n, 69.55% o
i 25.94% n, 74.06% o j 85.6% c, 14.4% h k 18.29% h, 81.71% c l 10.3% c, 89.87% cl

Explanation:

Step1: Assume 100g of the compound

This allows us to consider the percentages as masses in grams. For example, for 40.68% C, 5.12% H, 54.20% O, we have 40.68g C, 5.12g H and 54.20g O.

Step2: Calculate the moles of each element

The molar mass of C is 12.01g/mol, H is 1.01g/mol and O is 16.00g/mol. For C: $n_{C}=\frac{40.68g}{12.01g/mol}\approx3.39mol$. For H: $n_{H}=\frac{5.12g}{1.01g/mol}\approx5.07mol$. For O: $n_{O}=\frac{54.20g}{16.00g/mol}\approx3.39mol$.

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles among them. Here, the smallest is 3.39mol. For C: $\frac{3.39mol}{3.39mol}=1$. For H: $\frac{5.07mol}{3.39mol}\approx1.5$. For O: $\frac{3.39mol}{3.39mol}=1$. Multiply all by 2 to get whole - numbers, so C:H:O = 2:3:2. The empirical formula sub - script input is 232.
We will do the same for all the cases:

Case b: 15.8% C, 84.2% S

Assume 100g of the compound. $n_{C}=\frac{15.8g}{12.01g/mol}\approx1.32mol$, $n_{S}=\frac{84.2g}{32.07g/mol}\approx2.63mol$. Mole - ratio: $\frac{1.32mol}{1.32mol}=1$, $\frac{2.63mol}{1.32mol}\approx2$. Empirical formula sub - script input is 12.

Case c: 43% C, 57% O

Assume 100g of the compound. $n_{C}=\frac{43g}{12.01g/mol}\approx3.58mol$, $n_{O}=\frac{57g}{16.00g/mol}\approx3.56mol$. Mole - ratio: $\frac{3.58mol}{3.56mol}\approx1$, $\frac{3.56mol}{3.56mol}=1$. Empirical formula sub - script input is 11.

Case d: 12.1% C, 16.2% O, 71.7% Cl

Assume 100g of the compound. $n_{C}=\frac{12.1g}{12.01g/mol}\approx1.01mol$, $n_{O}=\frac{16.2g}{16.00g/mol}\approx1.01mol$, $n_{Cl}=\frac{71.7g}{35.45g/mol}\approx2.02mol$. Mole - ratio: $\frac{1.01mol}{1.01mol}=1$, $\frac{1.01mol}{1.01mol}=1$, $\frac{2.02mol}{1.01mol}=2$. Empirical formula sub - script input is 112.

Case e: 50.04% C, 5.59% H, 44.37% O

Assume 100g of the compound. $n_{C}=\frac{50.04g}{12.01g/mol}\approx4.17mol$, $n_{H}=\frac{5.59g}{1.01g/mol}\approx5.53mol$, $n_{O}=\frac{44.37g}{16.00g/mol}\approx2.77mol$. Divide by 2.77mol: For C: $\frac{4.17mol}{2.77mol}\approx1.5$, For H: $\frac{5.53mol}{2.77mol}\approx2$, For O: $\frac{2.77mol}{2.77mol}=1$. Multiply by 2, empirical formula sub - script input is 342.

Case f: 40.0% C, 6.7% H, 53.3% O

Assume 100g of the compound. $n_{C}=\frac{40.0g}{12.01g/mol}\approx3.33mol$, $n_{H}=\frac{6.7g}{1.01g/mol}\approx6.63mol$, $n_{O}=\frac{53.3g}{16.00g/mol}\approx3.33mol$. Mole - ratio: $\frac{3.33mol}{3.33mol}=1$, $\frac{6.63mol}{3.33mol}\approx2$, $\frac{3.33mol}{3.33mol}=1$. Empirical formula sub - script input is 121.

Case g: 45.8% S, 54.2% F

Assume 100g of the compound. $n_{S}=\frac{45.8g}{32.07g/mol}\approx1.43mol$, $n_{F}=\frac{54.2g}{19.00g/mol}\approx2.85mol$. Mole - ratio: $\frac{1.43mol}{1.43mol}=1$, $\frac{2.85mol}{1.43mol}\approx2$. Empirical formula sub - script input is 12.

Case h: 30.45% N, 69.55% O

Assume 100g of the compound. $n_{N}=\frac{30.45g}{14.01g/mol}\approx2.17mol$, $n_{O}=\frac{69.55g}{16.00g/mol}\approx4.35mol$. Mole - ratio: $\frac{2.17mol}{2.17mol}=1$, $\frac{4.35mol}{2.17mol}\approx2$. Empirical formula sub - script input is 12.

Case i: 25.94% N, 74.06% O

Assume 100g of the compound. $n_{N}=\frac{25.94g}{14.01g/mol}\approx1.85mol$, $n_{O}=\frac{74.06g}{16.00g/mol}\approx4.63mol$. Mole - ratio: $\frac{1.85mol}{1.85mol}=1$, $\frac{4.63mol}{1.85mol}\approx2.5$. Multiply by 2, empirical formula sub - script input is 25.

Case j: 85.6% C, 14.4% H

Assume 100g of the compound. $n_{C}=\frac{85.6g}{12.01g/mol}\approx7.13mol$, $n_{H}=\frac{14.4g}{1.01g/mol}\approx14.26mol$. Mole - ratio: $\frac{7.13mol}{7.13mol}=1$, $\frac{14.26mol}{7.13mol}=2$. Empirical formula sub - script input is 12.

Case k: 18.29% H, 81.71% C

Assume 100g of the compound. $n_{H}=\frac{18.29g}{1.01g/mol}\approx18.11mol$, $n_{C}=\frac{81.71g}{12.01g/mol}\approx6.80mol$. Mole - ratio: $\frac{6.80mol}{6.80mol}=1$, $\frac{18.11mol}{6.80mol}\approx2.66$. Multiply by 3, empirical formula sub - script input is 38.

Case l: 10.3% C, 89.87% Cl

Assume 100g of the compound. $n_{C}=\frac{10.3g}{12.01g/mol}\approx0.86mol$, $n_{Cl}=\frac{89.87g}{35.45g/mol}\approx2.54mol$. Mole - ratio: $\frac{0.86mol}{0.86mol}=1$, $\frac{2.54mol}{0.86mol}\approx3$. Empirical formula sub - script input is 13.

Answer:

a. 232
b. 12
c. 11
d. 112
e. 342
f. 121
g. 12
h. 12
i. 25
j. 12
k. 38
l. 13