The atomic weight of helium is 4.0026 u. What is the mass of a helium sample which contains 0.427 moles of He gas? Select one: a. 0.427 g b. 0.107 g c. 1.71 g d. 2.57 g e. 9.37 g
The Correct Answer and Explanation is:
Step 1: Identify Given Data
- Moles of He gas: 0.427 mol
- Atomic mass (molar mass) of helium (He): 4.0026 g/mol
Step 2: Apply the Formula
mass=0.427 mol×4.0026 gmol=1.7091 g\text{mass} = 0.427 \, \text{mol} \times 4.0026 \, \frac{\text{g}}{\text{mol}} = 1.7091 \, \text{g}mass=0.427mol×4.0026molg=1.7091g
Rounding to three significant figures (to match the input data):mass≈1.71 g\text{mass} \approx \boxed{1.71 \, \text{g}}mass≈1.71g
Correct Answer: c. 1.71 g
Textbook-Style Explanation
In chemical calculations, one of the most fundamental relationships involves converting between the amount of substance (in moles) and mass (in grams). This relationship is governed by the molar mass of the substance, which is numerically equal to its atomic or molecular weight expressed in grams per mole (g/mol). In this example, we are given a sample of helium (He), a noble gas, and asked to determine its mass when the number of moles is known.
The atomic weight of helium is 4.0026 atomic mass units (u), and therefore, one mole of helium atoms weighs 4.0026 grams. This is the molar mass. The molar mass acts as a conversion factor between moles and grams. To find the mass, we multiply the number of moles by the molar mass.
Given 0.427 moles of helium, we perform the calculation as follows:mass=0.427 mol×4.0026 g/mol=1.7091 g\text{mass} = 0.427 \, \text{mol} \times 4.0026 \, \text{g/mol} = 1.7091 \, \text{g}mass=0.427mol×4.0026g/mol=1.7091g
We round the result to three significant figures to match the precision of the original data, giving us a final answer of 1.71 grams.
This problem reinforces the concept that knowing the amount of a substance in moles, along with its molar mass, allows for straightforward conversion to mass, which is essential in chemical reactions and laboratory measurements.
Answer: c. 1.71 g ✅
