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
[tex]\displaystyle 675 \ g \ CBr_4(\frac{1 \ mol \ CBr_4}{331.61 \ g \ CBr_4}) = 2.03552 \ mol \ CBr_4[/tex]
Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
2.03552 mol CBr₄ ≈ 2.04 mol CBr₄
Answer:
[tex]\boxed {\boxed {\sf About \ 2.04 \ moles \ of \ CBr_4}}[/tex]
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.
[tex]\frac{331.611 \ g \ CBr_4}{1 \ mol \ CBr_4 }[/tex]
Multiply by the given number of grams.
[tex]675 \ g \ CBr_4 *\frac{331.611 \ g \ CBr_4}{1 \ mol \ CBr_4 }[/tex]
Flip the fraction so the grams of CBr₄ will cancel.
[tex]675 \ g \ CBr_4 *\frac{1 \ mol \ CBr_4 }{331.611 \ g \ CBr_4}[/tex]= [tex]675 \ *\frac{1 \ mol \ CBr_4 }{331.611 \ }[/tex]
[tex]\frac{675 \ mol \ CBr_4 }{331.611 \ }[/tex] = [tex]2.03551752 \ mol \ CBr_4[/tex]
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.
[tex]2.03551752 \ mol \ CBr_4[/tex]
The 5 in the thousandth place tells us to round the 3 to a 4.
[tex]\approx 2.04 \ mol \ CBr_4[/tex]
There are about 2.04 moles of carbon tetrabromide in 675 grams.
