Question

how many formula units are there in 216g of Ba(NO3)2?

Answer 1

Answer:

4.98 × 10²³ formula units Ba(NO₃)₂

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
• Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

• Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 216 g Ba(NO₃)₂

[Solve] formula units Ba(NO₃)₂

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Ba - 137.33 g/mol

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

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

Molar Mass of Ba(NO₃)₂ - 137.33 + 2[14.01 + 3(16.00)] = 261.35 g/mol

Step 3: Convert

1. [DA] Set up:                                                                                                      $\displaystyle 216 \ g \ Ba(NO_3)_2(\frac{1 \ mol \ Ba(NO_3)_2}{261.35 \ g \ Ba(NO_3)_2})(\frac{6.022 \cdot 10^{23} \ formula \ units \ Ba(NO_3)_2}{1 \ mol \ Ba(NO_3)_2})$
2. [DA] Divide/Multiply [Cancel out units]:                                                          $\displaystyle 4.97705 \cdot 10^{23} \ formula \ units \ Ba(NO_3)_2$

Step 4: Check

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

4.97705 × 10²³ formula units Ba(NO₃)₂ ≈ 4.98 × 10²³ formula units Ba(NO₃)₂

Answer 2

Answer:

Explanation:

Given 216g Ba(NO₃)₂ = 216g/261.34g/mol = 0.827mol

0.827mol = 0.827mol x 6.023 x 10²³molecules/mol

= 4.978066886 x 10²³ molecules (calc. ans.)

≅ 4.98 x 10²³ molecules (3 sig. figs. based on given 216g Ba(NO₂)₃)