Midterm 1

General Chemistry 1

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Question 1

Hydrochloric acid and sodium sulfide are mixed and totally react to form hydrogen sulfide and sodium chloride. Write the chemical equation of this reaction. How many grams of hydrogen sulfide are produced from 5.0 mol of both hydrochloric acid and sodium sulfide?

Solution

2 HCl(aq) + Na₂S(aq) → H₂S(g) + 2 NaCl(aq) We have to determine the limiting reactant: - if HCl is the limiting reactant: at t_final, n_HCl – 2x₁ = 0 ⇒ x₁ = 2.5 mol - if Na₂S is the limiting reactant: at t_final, n_Na2S – x₂ = 0 ⇒ x₂= 5.0 mol x₁ < x₂ so HCl is the limiting reactant and n_H2S = x₁ = 2.5 mol n_H2S =

\frac{m_{H 2 S}}{M_{H 2 S}}

⇒ m_H2S = n_H2S x M_H2S = 85 g

Question 2

A sample of an amino acid (M = 225.2 g.mol^-1) contains 1.56 g of C, 0.333 g of H, 2.08 g of O, and 0.910 g of N.

What are the empirical and molecular formulas of this amino acid?

Solution

Empirical Formula:

Determine the number of atoms for each element:

n_C = m_C / M_C = 0.130 mol

n_H = m_H / M_H = 0.330 mol

n_O = 0.130 mol

n_N = 0.0650 mol

⇒ n_N= 2n_C = 2n_O = 5n_H

The empirical formula of this amino acid is: C₂H₅O₂N

Molecular Formula:

The molar mass of C₂H₅O₂Nis 75.05 g.mol^-1

The molar mass of this amino acid is 225.2 g.mol^-1

225.2/75.05 = 3, the molecular formula of vitamin C is C₆H₁₅O₆N₃

Question 3

Give the name or the formula of the following compounds and specify if they are ionic or covalent:

Dicarbon tetroxide

Phosphoric acid

Calcium chlorite

Copper(I) bromide

Pb(ClO₃)₄

SF₆

AgI

Solution

Dicarbon tetroxide: C₂O₄ – covalent compound

Phosphoric acid : H₃PO₄ – covalent compound

Calcium chlorite: Ca(ClO₂)₂ – ionic compound

Copper(I) bromide: CuBr – ionic compound

HF : hydrofluoric acid – covalent compound

Pb(ClO₃)₄: Lead(IV) chlorate – ionic compound

SF₆: sulfur hexafluoride – covalent compound

AgI: silver iodide - ionic compound

Question 4

Give the number of protons, the number of neutrons, the charge and the atomic symbol of the following elements:

oxygen with 10 electrons (A = 16)

aluminum with 13 electrons (A = 27)

tungsten with 74 electrons (A = 183)

tin with 46 electrons (A = 120)

Solution

oxygen with 10 electrons (A = 16): 8 protons, 8 neutrons, - 2, ${}_{8}^{16}O^{2 -}$

aluminum with 13 electrons (A = 27): 13 protons, 14 neutrons, 0, ${}_{13}^{27}{Al}$

tungsten with 74 electrons (A = 183): 74 protons, 109 neutrons, 0, ${}_{74}^{183}W$

tin with 46 electrons (A = 120): 50 protons, 70 neutrons, +4, ${}_{50}^{120}{Sn}^{4 +}$

Question 5

Selenium atom has 6 main isotopes:

⁷⁴Se (0.860%), ⁷⁶Se (9.23%), ⁷⁷Se (7.60%), ⁷⁸Se (23.7%), ⁸⁰Se(49.8%) and ⁸²Se (8.82%).

What is the average atomic mass of selenium?

On another planet, the selenium atom has only 2 main isotopes: ⁷⁷Se and ⁸⁰Se.

Its atomic mass is the same as that on Earth.

Determine the relative percentage of these two isotopes.

Solution

Average atomic mass of selenium on Earth:

74 x 0.00860 + 76 x 0.0923 + 77 x 0.0760 + 78 x 0.237 + 80 x 0.498 + 82 x 0.0882 = 79.1 amu

Relative percentage of ⁷⁷Se and ⁸⁰Se on the other planet:

% of ⁷⁷Se = X%

% of ⁸⁰Se = Y% = (100-X)%

79.1 amu = $\frac{77 X}{100}$ + $\frac{80 Y}{100}$ = $\frac{77 X}{100}$ + $\frac{80 (100 - X)}{100}$ ⇒ X = 30

% of ⁷⁷Se = 30.0%

% of ⁸⁰Se = 70.0%

Question 6

Sodium cyanide is very useful in inorganic chemistry since the cyanide anion has a high affinity for metals, especially with gold.

What is the molar mass of this compound?

Calculate the percent composition by mass for each element in sodium cyanide.

How many carbon atoms are found in 1.2 pounds of solution if the solution is 5.0% sodium cyanide by mass?

(1.0 pound = 4.5 x 10² g)

Solution

Sodium cyanide: NaCN ⇒ M_NaCN = M_Na + M_C + M_N = 23.0 + 12.0 + 14.0 = 49.0 g.mol^-1

% composition by mass:

% Na = M_Na/ M_NaCN x 100 = 23.0 / 49.0 x 100 = 46.9%

% C = 24.5%

% N = 28.6 %

Number of carbon atoms N:

n_C = N / N_A

Because we have only one carbon atom in the molecule NaCN: n_C = n_NaCN = m_NaCN/ M_NaCN

m_NaCN = 1.2 pounds x (4.5 x 10²) g / 1 pound x 0.050 = 2.7 x 10 g

N = N_A x n_C = N_A x n_NaCN = N_A x m_NaCN/ M_NaCN = 6.022 x 10²³ x 2.7 x 10 / 49.0 = 3.3 x 10²³ atoms

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