Question

4. How many moles of gold (Au) are in a pure gold nugget having a mass of 25.0 grams.

Answer 1

Answer:

0.127 mol Au

General Formulas and Concepts:

Math

Order of Operations: BPEMDAS

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

• Left to Right

Chemistry

Atomic Structure

• Reading a Periodic Table
• Moles

Stoichiometry

• Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 25.0 g Au

[Solve] moles Au

Step 2: Identify Conversions

[PT] Molar Mass of Au - 196.97 g/mol

Step 3: Convert

1. [DA] Set up:                                                                                                     $\displaystyle 25.0 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})$
2. [DA] Multiply/Divide [Cancel out units:                                                          $\displaystyle 0.126923 \ mol \ Au$

Step 4: Check

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

0.126923 mol Au ≈ 0.127 mol Au

Answer 2

Answer:

$\boxed {\boxed {\sf 0.127 \ mol \ Au}}$

Explanation:

To convert from mass to moles, the molar mass must be used. This can be found on the Periodic Table.

• Gold (Au): 196.96657 g/mol

Use this number as a ratio.

$\frac {196.96657 \ g\ Au }{1 \ mol \ Au}$

Multiply by the given number of grams.

$25.0 \ g \ Au*\frac {196.96657 \ g\ Au }{1 \ mol \ Au}$

Flip the fraction so the grams of gold will cancel.

$25.0 \ g \ Au*\frac {1 \ mol \ Au}{196.96657 \ g\ Au }$

$25.0*\frac {1 \ mol \ Au}{196.96657 }$

$\frac {25.0 \ mol \ Au}{196.96657 }$

$0.1269250919 \ mol \ Au$

The original measurement has 3 significant figures, so the answer must have the same. For the number we calculated that is the thousandth place. The 9 in the ten thousandth place tells us to round the 6 to a 7.

$0.127 \ mol \ Au$

25.0 grams of gold is about 0.127 moles of gold.