Question

675 g of carbon tetrabromide is equivalent to how many

Answer 1

Answer:

$\boxed {\boxed {\sf About \ 2.04 \ moles \ of \ CBr_4}}$

Explanation:

1. Define Formula

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

• Carbon (C)+ 4 Bromine (Br) = CBr₄

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.

• Carbon (C): 12.011 g/mol
• Bromine (Br): 79.90 g/mol

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.

$\frac{331.611 \ g \ CBr_4}{1 \ mol \ CBr_4 }$

Multiply by the given number of grams.

$675 \ g \ CBr_4 *\frac{331.611 \ g \ CBr_4}{1 \ mol \ CBr_4 }$

Flip the fraction so the grams of CBr₄ will cancel.

$675 \ g \ CBr_4 *\frac{1 \ mol \ CBr_4 }{331.611 \ g \ CBr_4}$= $675 \ *\frac{1 \ mol \ CBr_4 }{331.611 \ }$

$\frac{675 \ mol \ CBr_4 }{331.611 \ }$ = $2.03551752 \ mol \ CBr_4$

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.

$2.03551752 \ mol \ CBr_4$

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

$\approx 2.04 \ mol \ CBr_4$

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

Answer 2

Answer:

2.04 mol CBr₄

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Chemistry

Organic

• Writing Organic Compounds
• Writing Covalent Compounds
• Organic Prefixes

Atomic Structure

• Reading a Periodic Table
• Using Dimensional Analysis

Explanation:

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

$\displaystyle 675 \ g \ CBr_4(\frac{1 \ mol \ CBr_4}{331.61 \ g \ CBr_4}) = 2.03552 \ mol \ CBr_4$

Step 4: Check

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

2.03552 mol CBr₄ ≈ 2.04 mol CBr₄