Chemistry, 06.03.2021 06:40 Welcomes

What is the mass in grams of 1.00 x 10^12 lead (Pb)
atoms?

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Answers

Answer from: jaylabazemore

3.44 × 10⁻¹⁰ g Pb

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Chemistry

Atomic Structure

Reading a Periodic TableAvogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional AnalysisExplanation:

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

[DA] Set up: $\displaystyle 1.00 \cdot 10^{12} \ atoms \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ atoms \ Pb})(\frac{207.2 \ g \ Pb}{1 \ mol \ Pb})$ [DA] Multiply/Divide [Cancel out units]: $\displaystyle 3.44072 \cdot 10^{-10} \ g \ Pb$

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 from: abbydamaso5048

$\boxed {\boxed {\sf 3.44 *10^{-10} \ g \ Pb}}$

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.

$\frac {6.022*10^{23} \ atoms \ Pb}{1 \ mol \ Pb}$

Multiply by the given number of atoms.

$1.00 *10^{12} \ atoms \ Pb *\frac {6.022*10^{23} \ atoms \ Pb}{1 \ mol \ Pb}$

Flip the fraction so the atoms of lead cancel.

$1.00 *10^{12} \ atoms \ Pb *\frac {1 \ mol \ Pb}{6.022*10^{23} \ atoms \ Pb}$

$1.00 *10^{12}*\frac {1 \ mol \ Pb}{6.022*10^{23}}$

$\frac {1.00 *10^{12}\ mol \ Pb}{6.022*10^{23}} = 1.66057788*10^{-12} \ mol \ Pb$

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.

$\frac {207.2 \ g\ Pb}{ 1 \ mol \ Pb}$

Multiply by the number of moles we calculated.

$1.66057788*10^{-12} \ mol \ Pb* \frac {207.2 \ g\ Pb}{ 1 \ mol \ Pb}$

$1.66057788*10^{-12}* \frac {207.2 \ g\ Pb}{ 1 }$

$1.66057788*10^{-12}* {207.2 \ g\ Pb}= 3.44071737*10^{-10} \ g \ Pb$

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.

$3.44 *10^{-10} \ g \ Pb$

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