Question

1. base your answer(s) to the following question(s) on the information below.

arsenic is often obtained by heating the ore arsenopyrite, feass. the decomposition of feass is represented by the balanced equation below.
$ce{feass(s) xrightarrow{heat} fes(s) + as(g)}$
in the solid phase, arsenic occurs in two forms. one form, yellow arsenic, has a density of $1.97\\ \text{g/cm}^3$ at stp. the other form, gray arsenic, has a density of $5.78\\ \text{g/cm}^3$ at stp. when arsenic is heated rapidly in air, arsenic(iii) oxide is formed.
although arsenic is toxic, it is needed by the human body in very small amounts. the body of a healthy human adult contains approximately 5 milligrams of arsenic.
calculate the percent composition by mass of arsenic in arsenopyrite. your response must include both a correct numerical setup and the calculated result.

Explanation:

Step1: Calcular masa molar de FeAsS

Masa molar de Fe = 55.85 g/mol, As = 74.92 g/mol, S = 32.07 g/mol.
$$\text{Masa molar de FeAsS} = 55.85 + 74.92 + 32.07 = 162.84\ \text{g/mol}$$

Step2: Establecer fórmula de % composición

La fórmula es $\frac{\text{Masa de As en FeAsS}}{\text{Masa molar de FeAsS}} \times 100\%$
$$\% \text{ As} = \frac{74.92}{162.84} \times 100\%$$

Step3: Calcular el valor numérico

Realizar la división y multiplicación.
$$\% \text{ As} \approx 0.4601 \times 100\% = 46.01\%$$

Answer:

El porcentaje de masa de arsénico en la arsenopirita es aproximadamente $\boldsymbol{46.0\%}$ (o 46.01% con mayor precisión), con la configuración numérica $\frac{74.92}{55.85 + 74.92 + 32.07} \times 100\%$.