Question

0.2636g of nickel (ni) metal burns with 0.1440g of sulfur (s) to give nickel sulfide compound. what is the empirical formula of the compound? (1 ni atom = 58.69amu; 1 sulfur atom = 32.07amu). 1) ni₂s 2) nis 3) nis₂ 4) ni₂s₃

Explanation:

Step1: Calculate moles of Ni

Moles of Ni = mass / molar mass = \( \frac{0.2636\,\text{g}}{58.69\,\text{g/mol}} \approx 0.00449\,\text{mol} \)

Step2: Calculate moles of S

Moles of S = mass / molar mass = \( \frac{0.1440\,\text{g}}{32.07\,\text{g/mol}} \approx 0.00449\,\text{mol} \)

Step3: Find mole ratio

Ratio of Ni to S = \( \frac{0.00449}{0.00449} : \frac{0.00449}{0.00449} = 1:1 \)

Answer:

1. NiS