# Midterm 1

General Chemistry 1

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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?

2 HCl(aq) + Na2S(aq) → H2S(g) + 2 NaCl(aq)

We have to determine the limiting reactant:

- if HCl is the limiting reactant: at tfinal,  nHCl – 2x1 = 0 ⇒ x1 = 2.5 mol

- if Na2S is the limiting reactant: at tfinal,  nNa2S – x2 = 0 ⇒ x2 = 5.0 mol

x1 < x2 so HCl is the limiting reactant and nH2S = x1 = 2.5 mol

nH2S = $\frac{{\mathrm{m}}_{\mathrm{H}2\mathrm{S}}}{{\mathrm{M}}_{\mathrm{H}2\mathrm{S}}}$ ⇒ mH2S = nH2S x MH2S = 85 g

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?

Empirical Formula:

Determine the number of atoms for each element:

nC = mC / MC = 0.130 mol

nH = mH / MH = 0.330 mol

nO = 0.130 mol

nN = 0.0650 mol

⇒ nN = 2nC = 2nO = 5nH

The empirical formula of this amino acid is: C2H5O2N

Molecular Formula:

The molar mass of C2H5O2N is 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

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

HF

Pb(ClO3)4

SF6

AgI

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

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)

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

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

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

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

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.

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 =  +  =  +  ⇒ X = 30

% of 77Se = 30.0%

% of 80Se = 70.0%

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)

Sodium cyanide: NaCN ⇒ MNaCN = MNa + MC + MN = 23.0 + 12.0 + 14.0 = 49.0 g.mol-1

% composition by mass:

% Na = MNa / MNaCN x 100 = 23.0 / 49.0 x 100 = 46.9%

% C = 24.5%

% N = 28.6 %

Number of carbon atoms N:

nC = N / NA

Because we have only one carbon atom in the molecule NaCN: nC = nNaCN = mNaCN / MNaCN

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

N = NA x nC = NA x nNaCN = NA x mNaCN / MNaCN = 6.022 x 1023 x 2.7 x 10 / 49.0 = 3.3 x 1023 atoms