675 g of carbon tetrabromide is equivalent to how many mol. Make sure to round your answer to the nearest WHOLE NUMBER

2.04 mol CBr₄

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Organic

Atomic Structure

Step 1: Define

675 g CBr₄

Step 2: Identify Conversions

Molar Mass of C - 12.01 g/mol

Molar Mass of Br - 79.90 g/mol

Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

2.03552 mol CBr₄ ≈ 2.04 mol CBr₄

Explanation:

1. Define Formula

The compound is carbon tetrabromide. The tetra indicates 4 atoms of bromine.

2. Find Molar Mass

There are 2 elements in this compound: carbon and bromine. Use the Periodic Table to find the molar masses of these elements.

The molar mass is based on the number of atoms in the compound. The compound has 1 atom of carbon and 4 atoms of bromine.

CBr₄= 1(12.011 g/mol) + 4(79.90 g/mol)

= 12.011 g/mol+319.60 g/mol = 331.611 g/mol

3. Convert Grams to Moles

We want to convert 675 grams to moles. We should use the molar mass as a fraction.

Multiply by the given number of grams.

Flip the fraction so the grams of CBr₄ will cancel.

=

=

4.Round

The original measurement, 675 grams has 3 significant figures (6,7 and 5), so our answer must have the same.

For the answer we found, 3 sig figs is the hundredth place.

The 5 in the thousandth place tells us to round the 3 to a 4.

There are about 2.04 moles of carbon tetrabromide in 675 grams.

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