Question

Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!

Convert 0.2502 grams of calcium nitrate to moles.

How many atoms are contained in 3.80 moles of magnesium?

Answer 1

a) 0.001525 mol Ca(NO₃)₂

b) 2.29 × 10²⁴ atoms Mg

General Formulas and Concepts:

Atomic Structure

• Reading a Periodic Table
• Reactions rxn
• Compounds

Step-by-step explanation:

a)

Calcium (Ca) has a +2 charge on the periodic table. Polyatomic ion for nitrate (NO₃) has a -1 charge. We need to balance the compound:

Ca(NO₃)₂

To find the molar mass of calcium nitrate, we must pull data from the PT.

Ca has a molar mass of 40.08 g/mol

N has a molar mass of 14.01 g/mol

O has a molar mass of 16.00 g/mol

Add these all together and multiply respectively to find the molar mass of the compound:

40.08 + 2[14.01 + 3(16.00)] = 164.1 g/mol

To find the amount of moles we have given grams, divide the grams by the molar mass:

(0.2502 g) / (164.1 g/mol) = 0.001525 mol Ca(NO₃)₂

We are given 4 sig figs and our final answer is in 4 sig figs.

b)

We need to know Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc. This converts moles into atoms, molecules, formula units, etc.

We are given 3.80 moles of Mg. To find the number of atoms, we multiply Avogadro's number:

3.80 mol Mg · (6.022 × 10²³ atoms Mg/ 1 mol Mg) = 2.28836 × 10²⁴ atoms Mg

Since we are given 3 sig figs, we must round:

2.28836 × 10²⁴ atoms Mg ≈ 2.29 × 10²⁴ atoms Mg

Answer 2

Solution given:

a.

we have

1 mole of calcium nitrate=164.088g

now

0.2502g of calcium nitrate=1/164.088*0.2502=

0.0015248 mole of calcium nitrate

0.2502 grams of calcium nitrate

=0.0015248moles.

b.

1 mole of magnesium=
`6.02214076 × 10^(23)`atoms

3.80 mole of magnesium=
`6.02214076 × 10^(23)×3.80=2.2884*10^(24)`atoms

`\bold{\underline{2.2884*10^(24)}}`atoms

are contained in 3.80 moles of magnesium.