The mole and the Avogadro constant

Solid sodium carbonate reacts with aqueous hydrochloric acid to form aqueous sodium chloride, carbon dioxide and water.

`Na_2CO_3 + 2HCl -> 2NaCl + CO_2 + H_2O`

a. Rewrite this equation to include state symbols

b. Calculate the number of moles of hydrochloric acid required to react exactly with 4.15 g of sodium carbonate.

(A_r values: C = 12.0, Na = 23.0, O = 16.0)

c. Define the term mole.

a. `Na_2CO_3 (s) + 2HCl (aq) -> 2NaCl (aq) + CO_2 (g) + H_2O (l)`

b. The molar mass of `Na_2CO_3 = 2 xx 23.0 + 12.0 + 3 xx 16.0 = 106 g.mol^(-1`

Number of moles = `"Mass" / "Molar mass" = 4.15 / 106 = 0.0391 "moles"`

Number of moles of HCl = 2 x number of moles of Na₂CO₃ = `2 xx 0.0391 = 0.0782 "moles"`

c. One mole of a substance is the amount of that substance that has the same number of specific particles (atoms, molecules or ions) as there are atoms in exactly 12 g of the carbon-12 isotope.

What is Avogadro constant?

The number of atoms in a mole of atoms is very large: `6.02 × 10^23` atoms. This number is called the Avogadro constant (or Avogadro number). The symbol for the Avogadro constant is L (the symbol N_A may also be used).

a. An aqueous solution of 25.0 cm³ sodium carbonate of concentration 0.0200 mol dm^–3 is titrated with hydrochloric acid. The volume of hydrochloric acid required to exactly react with the sodium carbonate is 12.50 cm³

i. Calculate the number of moles of sodium carbonate present in the solution of sodium carbonate.

ii. Calculate the concentration of the hydrochloric acid.

b. How many moles of carbon dioxide are produced when 0.2 mol of sodium carbonate reacts with excess hydrochloric acid?

c. Calculate the volume of this number of moles of carbon dioxide at r.t.p. (1 mol of gas occupies 24 dm³ at r.t.p.)

As we know:

`Na_2CO_3 (s) + 2HCl (aq) -> 2NaCl (aq) + CO_2 (g) + H_2O (l)`

a. i. Volume in dm³ = `25.0 / 1000 = 0.025 dm^3`

Number of moles of sodium carbonate = `0.025 xx 0.0200 = 0.0005 "moles"`

ii. Number of moles of hydrochloric acid = `0.0005 xx 2 = 0.0010 "moles"`

b. Excess hydrochloric means we use a large amount of acid to react with sodium carbonate completely.

Thus, number of moles of sodium carbonate = number of moles of carbon dioxide = 0.2 mol.

c. The volume of carbon dioxide at r.t.p = `0.2 xx 24 = 4.8 dm^3`

When sodium reacts with titanium chloride (TiCl₄), sodium chloride (NaCl) and titanium (Ti) are produced. a Write the balanced symbol equation for the reaction.

a. Write the balanced symbol equation for the reaction.

b. What mass of titanium is produced from 380 g of titanium chloride? Give your answer to 3 significant figures.

(A_r values: Ti = 47.9, Cl = 35.5)

a. `4 Na + TiCl_4 -> 4 NaCl + Ti`

b. Molar mass of `TiCl₄` = (1 × atomic mass of Ti) + (4 × atomic mass of Cl) 

Molar mass of `TiCl₄` = (1 × 47.9 g/mol) + (4 × 35.5 g/mol) 

Molar mass of `TiCl₄` = 189.900 g/mol

Thus,

Moles of `TiCl₄` = `("Given mass of TiCl₄")/ ("Molar mass of TiCl₄")` 

Moles of `TiCl₄` = `380 / 189.900 = 2.001 "moles"`

From the equation, we see that 1 mole of `TiCl₄` produces 1 mole of titanium.

Therefore, mass of titanium = Moles of titanium × Molar mass of titanium = `2.001 × 47.9 = 95.848 g`

Give all answers to 3 significant figures. Ammonium nitrate decomposes on heating to give nitrogen(I) oxide and water as follows:

`NH_4NO_3(s) -> N_2O(g) + 2H_2O(l)`

a. What is the formula mass of ammonium nitrate?

b. How many moles of ammonium nitrate are present in 0.800 g of the solid?

c. What volume of N₂O gas would be produced from this mass of ammonium nitrate?

a. The formula of ammonium nitrate is NH₄NO₃. The atomic masses of each element: 

Nitrogen (N): 2 × 14.0 g/mol = 28.0 g/mol 

Hydrogen (H): 4 × 1.0 g/mol = 4.0 g/mol 

Oxygen (O): 3 × 16.0 g/mol = 48.0 g/mol 

Thus, total formula mass = 28.0 + 4.0 + 48.0 = 80.0 g/mol.

b. Number of moles of ammonium nitrate = `0.800 / 80.0` = 0.0100 moles

c. When a question is about determining the volume of gas but not referring to temperature and pressure condition, we can assume that condition at Standard Temperature and Pressure (STP) where 1 mole of gas occupies 22.4 liters or 22.4 dm³.

The moles of N₂O gas = The moles of NH₄NO₃ = 0.0100 moles

Thus, the volume of N₂O gas = moles of N2O gas x 22.4 = 0.0100 x 22.4 = 0.224 L

Calcium oxide reacts with hydrochloric acid according to the equation:

`CaO + 2HCl -> CaCl_2 + H2O`

a. What mass of calcium chloride is formed when 28.05 g of calcium oxide reacts with excess hydrochloric acid?

b. What mass of hydrochloric acid reacts with 28.05 g of calcium oxide?

c. What mass of water is produced?

a. Molar mass of CaO = 40.08 g/mol + 16.00 g/mol = 56.08 g/mol 

Moles of CaO = `28.05 / 56.08` = 0.499 moles

From the equation, 1 mole of CaO produces 1 mole of CaCl₂

Thus, mass of calcium chloride = `(40.08 + 2 xx 35.5) xx 0.499 = 55.429 g`

b. From the equation, 1 mole of CaO produces 2 mole of HCl

Thus mass of hydrochloric acid = `(1.0 + 35.5) xx 2 xx 0.499 = 36.427 g`

c. Since 1 mole of CaO produces 1 mole of H₂O, the moles of water produced will also be approximately 0.499 mol.

Thus, mass of water = `(2 xx 1.0 + 16.0) xx 0.499 = 8.982 g`.

Which would contain 9.03×10²³ oxygen atoms?

A. 0.25 mol aluminium oxide 

B. 0.75 mol sulfur dioxide 

C. 1.5 mol sulfur trioxide 

D. 3.0 mol water

The answer is B.

Calculating the number of oxygen atoms in each given mole of the compounds to know which option would contain 9.03×10²³ oxygen atoms

Al₂O₃: 1 mole of Al₂O₃ ​contains 3 moles of oxygen atoms. Thus, 0.25 moles of Al₂O₃ ​ contains `0.25×3=0.75` moles of oxygen atoms. Number of oxygen atoms = c`0.75×6.022×10^23 =4.5165×10^23 "atoms"`.

SO₂ : 1 mole of SO₂ ​ contains 2 moles of oxygen atoms. Thus, 0.75 moles of SO₂ ​contains `0.75×2=1.5` moles of oxygen atoms. Number of oxygen atoms = `1.5×6.022×10^23 =9.033×10^23 "atoms"`.

SO₃: 1 mole of SO₃ ​ contains 3 moles of oxygen atoms. 1.5 moles of SO₃ ​ contains `1.5×3=4.5` moles of oxygen atoms. Number of oxygen atoms = `4.5×6.022×10^23 =2.71×10^24 "atoms"`.

H₂O: 1 mole of H₂O contains 1 mole of oxygen atoms. 3.0 moles of H₂O contains `3.0×1=3.0` moles of oxygen atoms. Number of oxygen atoms = `3.0×6.022×10^23 =1.8066×10^24 "atoms"`.

A 3.7 g sample of copper(II) carbonate is added to 25 cm³ of 2.0 mol dm⁻³ hydrochloric acid. Which volume of gas is produced under room conditions?

A. 0.6 dm³

B. 0.72 dm³

C. 1.20 dm³

D. 2.40 dm³

The answer is A.

`CuCO_3 (s)+ 2HCl (aq) → CuCl_2 (aq) + CO_2 (g) + H_2O (l)`

Molecular weight of CuCO3 ​ = 64.0 + 12.0 + 3 x 16.0 = 124.0 g 

Number of moles in 3.7 g of CuCO₃ ​ = `3.7 / 124.0` ​ = 0.03 mol 

Also, 2 mol dm⁻³ HCl means that number of moles HCl in 1 dm³ volume is 2. Therefore, number of moles of HCl in 25 cm³ = `25/1000 ​ × 2 = 0.05 mol`.

Here HCl is limiting reagent. Therefore, from the above reaction, volume of gas produced under room temperature when 2 moles of HCl reacts = 24 dm³.

Volume of gas produced when 0.05 moles of HCl reacts = `24/2​ × 0.05= 0.60 dm^3`

Diamond is a pure form of carbon. The mass of a diamond can be measured in carats. One carat is 0.200 g of carbon. Which expression gives the number of carats that contain 6.02×10²³ carbon atoms?

A. `0.200×12.0`

B. `0.200 / 12.0`

C. `12.0 / 0.200`

D. `0.200 / (6.02 xx 10^23) xx 12.0`

The answer is C.

The number of moles in 12 g carbon is 1 mol. 

The number of carbon atom in 1 mol of carbon is 6.022×10²³

The number of carats in 12 g of carbon `12 / 0.2`

Which contains the largest number of hydrogen atoms?

A. 0.10 mol of pentane

B. 0.20 mol of but-2-ene

C. 1.00 mol of hydrogen molecules

D. 6.02×10²³ hydrogen atoms

The answer is C.

The number of hydrogen atom in 0.1 mol of pentane (C_5​H₁₂)​ = `12 xx 0.10 xx 6.02 xx 10^23 = 7.2 xx 10^23`

The number of hydrogen atom in 0.2 mol of but−2−ene (C_4​H₈​) = `8 xx 0.2 xx 6.02 xx 10^23 = 9.6 xx 10^23`

In 1 mol of hydrogen molecule, the number of hydrogen atom = `2 xx 1 xx 6.02 xx 10^23​ = 12.04 xx 10^23`