Question

Determine the number of moles of the compound and determine the number of each type of atom in each of

Answer 1

Explanation:

Moles is denoted by given mass divided by the molecular mass ,  

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

From the question ,  

( a ) potassium bromide, KBr

As we know , the molecular mass of KBr = 119 g/mol

from the question ,

given mass of KBr = 2.12 g

using , the above formula to calculate the moles ,

n = w / m  = 2.12 g / 119 g/mol = 0.0178 mol.

The moles of the corresponding atoms are -

moles of K = 0.0178 mol

moles of Br = 0.0178 mol.

(b) phosphoric acid, H₃PO₄

As we know , the molecular mass of H₃PO₄ = 98 g/mol

from the question ,

given mass of  H₃PO₄ = 0.1488 g

using , the above formula to calculate the moles ,

n = w / m  = 0.1488 g  / 98 g/mol = 0.0015 mol.

The moles of the corresponding atoms are -

moles of H = 3 * 0.0015 = 0.0045 mol

moles of P = 0.0015 mol

moles of 0 = 4* 0.0015 = 0.0060 mol

(c) calcium carbonate, CaCO₃

As we know , the molecular mass of CaCO₃ = 100 g/mol

from the question ,

given mass of  CaCO₃ =  23 kg

Since , 1 kg = 1000 g

given mass of CaCO₃ =  23 * 1000 = 23000 g

using , the above formula to calculate the moles ,

n = w / m  = 23000 g  / 100 g/mol = 230 mol .

The moles of the corresponding atoms are -

moles of Ca = 230 mol

moles of C = 230 mol

moles of O = 3 * 230 = 690 mol

(d) aluminium sulfate, Al₂(SO₄)₃

As we know , the molecular mass of Al₂(SO₄)₃ = 342 g/mol

from the question ,

given mass of Al₂(SO₄)₃ = 78.452 g

using , the above formula to calculate the moles ,

n = w / m  = 78.452 g / 342 g/mol = 0.229 mol.

The moles of the corresponding atoms are -

moles of Al = 2 * 0.229 = 0.458 mol

moles of S = 3 * 0.229 = 0.687 mol

moles of O = 12 * 0.229 = 2.748 mol.

(e) Caffeine, C₈H₁₀N₄O₂

As we know , the molecular mass of C₈H₁₀N₄O₂ = 194 g/mol

from the question ,

given mass of C₈H₁₀N₄O₂ = 0.1250 mg

Since , 1 mg = 1 / 1000 g

given mass of C₈H₁₀N₄O₂ = 0.1250 / 1000 g = 0.000125 g

using , the above formula to calculate the moles ,

n = w / m  = 0.000125 g  / 194  g/mol = 6.44 * 10 ⁻⁷ mol.

The moles of the corresponding atoms are -

moles of C = 8 * 6.44 * 10 ⁻⁷ mol. = 51.52  * 10 ⁻⁷ mol.

moles of H = 10 * 6.44 * 10 ⁻⁷ mol. = 64.4  * 10 ⁻⁷ mol.

moles of N = 4 * 6.44 * 10 ⁻⁷ mol. = 25.76  * 10 ⁻⁷ mol.

moles of O = 2 * 6.44 * 10 ⁻⁷ mol. = 12.88  * 10 ⁻⁷ mol.

Answer 2

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles.

• For A: 2.12 g of potassium bromide

Given mass of KBr = 2.12 g

Molar mass of KBr = 119 g/mol

Putting values in equation 1, we get:

$\text{Moles of KBr}=\frac{2.12g}{119g/mol}=0.0178mol$

1 mole of potassium bromide contains 1 mole of potassium element and 1 mole of bromide element.

So, 0.0178 moles of potassium bromide will contain $(1\times 0.0178\times 6.022\times 10^{23})=1.07\times 10^{22}$ number of potassium atoms and

• For B: 0.1488 g of phosphoric acid

Given mass of phosphoric acid = 0.1488 g

Molar mass of phosphoric acid = 98 g/mol

Putting values in equation 1, we get:

$\text{Moles of phosphoric acid}=\frac{0.1488g}{98g/mol}=0.00152mol$

1 mole of phosphoric acid contains 3 moles of hydrogen element, 1 mole of phosphorus element and 4 moles of oxygen element.

So, 0.00152 moles of phosphoric acid will contain $(3\times 0.00152\times 6.022\times 10^{23})=2.75\times 10^{21}$ number of hydrogen atoms, $(1\times 0.00152\times 6.022\times 10^{23})=9.15\times 10^{20}$ of phosphorus atoms and $(4\times 0.00152\times 6.022\times 10^{23})=3.66\times 10^{21}$ number of oxygen atoms.

• For C: 23 kg of calcium carbonate

Given mass of calcium carbonate = 23 kg = 23000 g   (Conversion factor: 1 kg = 1000 g)

Molar mass of calcium carbonate = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of calcium carbonate}=\frac{23000g}{100g/mol}=230mol$

1 mole of calcium carbonate contains 1 mole of calcium element, 1 mole of carbon element and 3 moles of oxygen element.

So, 230 moles of calcium carbonate will contain $(1\times 230\times 6.022\times 10^{23})=1.38\times 10^{26}$ number of calcium atoms,

• For D: 78.452 g of aluminium sulfate

Given mass of aluminium sulfate = 78.452 g

Molar mass of aluminium sulfate = 342.15 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium sulfate}=\frac{78.452g}{342.15g/mol}=0.229mol$

1 mole of aluminium sulfate contains 2 moles of aluminium element, 3 moles of sulfur element and 12 moles of oxygen element.

So, 0.229 moles of aluminium sulfate will contain $(2\times 0.229\times 6.022\times 10^{23})=2.76\times 10^{23}$ number of aluminium atoms, $(3\times 0.229\times 6.022\times 10^{23})=4.14\times 10^{23}$ of sulfur atoms and $(12\times 0.229\times 6.022\times 10^{23})=1.65\times 10^{24}$ number of oxygen atoms.

• For E: 0.1250 mg of caffeine

Given mass of caffeine = 0.1250 mg = $0.125\times 10^{-3}$ g   (Conversion factor: 1 g = 1000 mg)

Molar mass of caffeine = 194.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of caffeine}=\frac{0.125\times 10^{-3}g}{194.2g/mol}=6.44\times 10^{-7}mol$

1 mole of caffeine contains 8 moles of carbon element, 10 moles of hydrogen element, 4 moles of nitrogen element and 2 moles of oxygen element.

So, $6.44\times 10^{-7}$ moles of caffeine will contain $(8\times 6.44\times 10^{-7}\times 6.022\times 10^{23})=3.10\times 10^{18}$ number of carbon atoms, $(10\times 6.44\times 10^{-7}\times 6.022\times 10^{23})=3.88\times 10^{18}$ of hydrogen atoms, $(4\times 6.44\times 10^{-7}\times 6.022\times 10^{23})=1.55\times 10^{18}$ of nitrogen atoms and $(2\times 6.44\times 10^{-7}\times 6.022\times 10^{23})=7.76\times 10^{17}$ number of oxygen atoms.