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Molecule Count in 0.334 g of C?H?: Ethane Analysis
Chapter 3, Problem 3.26(choose chapter or problem)
How many molecules of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) are present in 0.334 g of \(\mathrm{C}_{2} \mathrm{H}_{6}\)?
Questions & Answers
QUESTION:
How many molecules of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) are present in 0.334 g of \(\mathrm{C}_{2} \mathrm{H}_{6}\)?
ANSWER:Step 1 of 2
Here, we are going to calculate the number of molecules of ethane in 0.334 g.
We know that,
The molar mass of ethane \(=2(12.01 \mathrm{~g})+6(1.008 \mathrm{~g})=30.07 \mathrm{~g} / \mathrm{mol}\)
1.0 mol of ethane \(=6.022 \times 10^{23}\) molecules of ethane
1.0 mol of ethane contains = 30.07 g
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Molecule Count in 0.334 g of C?H?: Ethane Analysis
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This instructional video guides you through the process of determining the number of molecules in a given sample. It illustrates the conversion from grams to moles using the molar mass of the substance and then to molecules using Avogadro's number, providing a practical example with ethane (C?H?) to calculate approximately 6.67 x 10²¹ molecules in a 0.334 g sample.