2.999 mol Br
Math
Pre-Algebra
Order of Operations: BPEMDAS
Chemistry
Atomic Structure
Step 1: Define
1.806 × 10²⁴ molecules Br
Step 2: Identify Conversions
Avogadro's Number
Step 3: Convert
[tex]\displaystyle 1.806 \cdot 10^{24} \ molecules \ Br(\frac{1 \ mol \ Br}{6.022 \cdot 10^{23} \ molecules \ Br} )[/tex] = 2.999 mol Br
Step 4: Check
We are given 4 sig figs. Follow sig fig rules and round.
Our final answer is already in 4 sig figs, so there is no need to round.
Answer:
[tex]\boxed {\boxed {\sf About \ 2.999 \ moles \ of \ bromine}}[/tex]
Explanation:
To convert from moles to molecules, we must use Avogadro's Number.
[tex]6.022*10^{23}[/tex]
This tells us the number of particles (atoms, molecules, ions, etc) in 1 mole. In this problem, the particles are molecules of bromine in 1 mole of bromine.
[tex]6.022*10^{23} \ molecules \ Br / 1 \ mol \ Br[/tex]
1. Convert from moles to molecules
Use Avogadro's number as a fraction or ratio.
[tex]\frac{6.022 *10^{23} \ molecules \ Br}{1 \ mol \ Br}[/tex]
Multiply this fraction by the given number of bromine molecules.
[tex]1.806 *10^{24} molecules \ Br *\frac{6.022 *10^{23} \ molecules \ Br}{1 \ mol \ Br}[/tex]
Flip the fraction so the molecules of bromine can cancel out.
[tex]1.806 *10^{24} \ molecules \ Br* \frac{1 \ mol \ Br}{6.022 *10^{23} \ molecules \ Br}[/tex]
[tex]1.806 *10^{24} * \frac{1 \ mol \ Br}{6.022 *10^{23}}[/tex]
Multiply and condense the expression into 1 fraction.
[tex]\frac{1.806 *10^{24} \ mol \ Br}{6.022 *10^{23} }[/tex]
Divide.
[tex]2.999003653 \ mol \ Br[/tex]
2. Round
The original measurement, 1.806*10^24 has 4 significant figures (1, 8, 0, and 6). We must round our answer to 4 sig figs. For the answer we found, that is the thousandth place.
[tex]2.999003653 \ mol \ Br[/tex]
The 0 in the ten-thousandth place tels us to leave the 9 in the thousandth place.
[tex]\approx 2.999 \ mol \ Br[/tex]
There are about 2.999 moles of bromine in 2.806 *10^23 molecules.
