Question

How many moles are contained 80.3 gram of sm(no3)3

Answer 1

Answer:

0.239 mol Sm(NO₃)₃

General Formulas and Concepts:

Math

Pre-Algebra

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] 80.3 g Sm(NO₃)₃

[Solve] moles Sm(NO₃)₃

Step 2: Identify Conversions

[PT] Molar Mass of Sm - 150.36 g/mol

[PT] Molar Mass of N - 14.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of Sm(NO₃)₃ - 150.36 + 3[14.01 + 3(16.00)] = 336.39 g/mol

Step 3: Convert

1. [DA] Set up:                                                                                                 $\displaystyle 80.3 \ g \ Sm(NO_3)_3(\frac{1 \ mol \ Sm(NO_3)_3}{336.39 \ g \ Sm(NO_3)_3})$
2. [DA] Divide [Cancel out units]:                                                                        $\displaystyle 0.238711 \ mol \ Sm(NO_3)_3$

Step 4: Check

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

0.238711 mol Sm(NO₃)₃ ≈ 0.239 mol Sm(NO₃)₃