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How many molecules are in a 189 g sample of carbon tetrabromide, CBr4?

User Chkdsk
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2 Answers

4 votes

Final answer:

To find the number of molecules in a 189 g sample of CBr4, calculate the molar mass, convert the mass to moles, and then use Avogadro's number to find there are approximately 3.427 × 1023 molecules.

Step-by-step explanation:

To determine the number of molecules in a 189 g sample of carbon tetrabromide, CBr4, you first need to calculate the molar mass of CBr4. This is done by adding the atomic masses of carbon (C) and four bromine (Br) atoms. Carbon has an atomic mass of 12 g/mol, and each bromine atom has an atomic mass of approximately 80 g/mol. Therefore, CBr4 has a molar mass of 12 g/mol + (4 × 80 g/mol) = 332 g/mol.

Next, you use Avogadro's number, which is 6.022 × 1023 molecules/mol, to find the number of molecules:

Number of moles = Mass of sample (g) / Molar mass (g/mol) = 189 g / 332 g/mol = 0.569 × 1 mol

Number of molecules = Number of moles × Avogadro's number = 0.569 × 6.022 × 1023 molecules/mol

After performing the calculation, you'll find that a 189 g sample of CBr4 contains approximately 3.427 × 1023 CBr4 molecules.

User Joel Quiles
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3.7k points
7 votes

Answer:

3.4321229*10^23 molecules.

Step-by-step explanation:

1 mole of CBr4 contains 6.0221409*10^23 molecules (Avogadro's Number).

The molar mass of this compound is 12.011 + 4(79.904)

= 331.627 g.

So the answer is (189/ 331.627) * 6.0221409*10^23

= 3.4321229*10^23 molecules.

User BlueSun
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3.3k points