Question

a scientist discusses the percent composition for iron oxide (fe₃o₄) shown in the following table.
what does this data show?

• there are 42.2 moles of o in fe₃o₄.
• for every 100 g of fe₃o₄, there are 27.8 g of o.
• for every 100 g of fe₃o₄, there are 66.7 g of fe.
• there are 67.8 moles remaining if o is removed.

1. choose the best answer.

which of the following could not represent the percent composition of compound xy?

• 45.72% x and 59.28% y
• 70.85% x and 29.15% y
• 18.64% x and 81.36% y
• 55.55% x and 44.45% y

1. choose the best answer.

which compound contains the highest percent composition of copper?

• cu₂s
• cus
• cus₂
• cu₂s₃

Explanation:

First Question (FeSO₄ Percent Composition Interpretation)

Step1: Define percent composition

Percent composition means, per 100 g of compound, the mass of each element equals its percentage value.

Step2: Match to options

For FeSO₄, 36.7% Fe means 36.7 g Fe per 100 g compound; 21.1% S means 21.1 g S per 100 g compound; 42.2% O means 42.2 g O per 100 g compound.

Step1: Check total percentage

Valid percent composition sums to 100%. Calculate total for each option:

• Option1: $45.72\% + 59.28\% = 105.00\%$
• Option2: $70.85\% + 29.15\% = 100.00\%$
• Option3: $18.55\% + 81.50\% = 100.05\%$
• Option4: $58.55\% + 41.45\% = 100.00\%$

Step2: Identify invalid total

A total over 100% cannot represent percent composition.

Step1: Calculate molar masses

• $\text{Cu}_2\text{O}$: $2\times63.55 + 16.00 = 143.10$ g/mol
• $\text{CuO}$: $63.55 + 16.00 = 79.55$ g/mol
• $\text{Cu}_2\text{S}$: $2\times63.55 + 32.07 = 159.17$ g/mol
• $\text{CuCl}_2$: $63.55 + 2\times35.45 = 134.45$ g/mol

Step2: Find Cu percent for each

• $\text{Cu}_2\text{O}$: $\frac{2\times63.55}{143.10} \times 100\% \approx 88.82\%$
• $\text{CuO}$: $\frac{63.55}{79.55} \times 100\% \approx 79.89\%$
• $\text{Cu}_2\text{S}$: $\frac{2\times63.55}{159.17} \times 100\% \approx 79.85\%$
• $\text{CuCl}_2$: $\frac{63.55}{134.45} \times 100\% \approx 47.27\%$

Step3: Compare percentages

Select the compound with the highest Cu percentage.

Answer:

For every 100 g of FeSO₄, there are 36.7 g of Fe.

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Second Question (Invalid Percent Composition)