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3. How many molecules of C9H304 are there in .1g of it?

User Mizux
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1 Answer

3 votes

Answer:

3 × 10²⁰ molecules C₉H₃O₄

Step-by-step explanation:

Step 1: Define

Avagadro's Number: 6.02 × 10²³ atoms, molecules, formula units, etc.

Molar Mass C - 12.01 g/mol

Molar Mass H - 1.01 g/mol

Molar Mass O - 16.00 g/mol

Molar Mass C₉H₃O₄ - 175.12 g/mol

Step 2: Use Dimensional Analysis


0.1 \hspace{3} g \hspace{3} C_9H_3O_4(\frac{1 \hspace{3} mol \hspace{3} C_9H_3O_4}{175.12 \hspace{3} g \hspace{3} C_9H_3O_4} )(\frac{6.02(10)^(23) \hspace{3} molecules \hspace{3} C_9H_3O_4}{1 \hspace{3} mol \hspace{3} C_9H_3O_4} ) = 3.43764 × 10²⁰ molecules C₉H₃O₄

Step 3: Simplify

We are give 1 sig fig.

3.43764 × 10²⁰ molecules C₉H₃O₄ ≈ 3 × 10²⁰ molecules C₉H₃O₄

User Vlz
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