3.44 × 10⁻¹⁰ g Pb
Math
Pre-Algebra
Order of Operations: BPEMDAS
Chemistry
Atomic Structure
Stoichiometry
Step 1: Define
[Given] 1.00 × 10¹² atoms Pb
[Solve] grams Pb
Step 2: Identify Conversions
Avogadro's Number
[PT] Molar Mass of Pb - 207.2 g/mol
Step 3: Convert
Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
3.44072 × 10⁻¹⁰ g Pb ≈ 3.44 × 10⁻¹⁰ g Pb
Answer:
[tex]\boxed {\boxed {\sf 3.44 *10^{-10} \ g \ Pb}}[/tex]
Explanation:
1. Atoms to Moles
Use Avogadro's Number to convert atoms to moles. This number: 6.022*10²³, tells us the number of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, the particles are atoms of lead.
[tex]\frac {6.022*10^{23} \ atoms \ Pb}{1 \ mol \ Pb}[/tex]
Multiply by the given number of atoms.
[tex]1.00 *10^{12} \ atoms \ Pb *\frac {6.022*10^{23} \ atoms \ Pb}{1 \ mol \ Pb}[/tex]
Flip the fraction so the atoms of lead cancel.
[tex]1.00 *10^{12} \ atoms \ Pb *\frac {1 \ mol \ Pb}{6.022*10^{23} \ atoms \ Pb}[/tex]
[tex]1.00 *10^{12}*\frac {1 \ mol \ Pb}{6.022*10^{23}}[/tex]
[tex]\frac {1.00 *10^{12}\ mol \ Pb}{6.022*10^{23}} = 1.66057788*10^{-12} \ mol \ Pb[/tex]
2. Moles to Grams
Use the molar mass to convert moles to grams. This can be found on the Periodic Table. For lead it is 207.2 grams per mole.
[tex]\frac {207.2 \ g\ Pb}{ 1 \ mol \ Pb}[/tex]
Multiply by the number of moles we calculated.
[tex]1.66057788*10^{-12} \ mol \ Pb* \frac {207.2 \ g\ Pb}{ 1 \ mol \ Pb}[/tex]
[tex]1.66057788*10^{-12}* \frac {207.2 \ g\ Pb}{ 1 }[/tex]
[tex]1.66057788*10^{-12}* {207.2 \ g\ Pb}= 3.44071737*10^{-10} \ g \ Pb[/tex]
3. Round
The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, it is the hundredth place. The 0 in the thousandth place tells us to leave the 4.
[tex]3.44 *10^{-10} \ g \ Pb[/tex]
