## Chapter 2 Practice Problems

** Example:** 2.154 Ammonia (NH

_{3}) and hydrazine (N

_{2}H

_{4}) are both compounds of nitrogen and hydrogen. Based on the law of multiple proportions, how many grams of hydrogen would you expect 2.34 g of nitrogen to comobine with to yield ammonia? To yield hydrazine?

The mass ratio between nitrogen and hydrogen in NH_{3}
is 14 g_{N} : 3(1 g_{H}).
$$\frac{{m}_{N}}{{m}_{N{H}_{3}}}=\frac{14{g}_{N}}{14{g}_{N}+3\left(1.008{g}_{H}\right)}$$
In the denominator, the mass of ammonia is given as one nitrogen mass plus
three hydrogens. This ratio persists no matter how much nitrogen we have.
So the ratio must be the same. Then, we can set up the following.
$$\frac{14{g}_{N}}{14{g}_{N}+3\left(1.008{g}_{H}\right)}=\frac{2.34{g}_{N}}{{m}_{N{H}_{3}}}$$
So, *m _{NH3}* is
then

*m*= 2.84544 g = 2.84 g.

_{NH3}

** Example:** 2.72 A sample of mercury with a mass of 114.0 g was
combined with 12.8 g of oxygen gas, and the resulting reaction gave 123.1 g
of mercury(II) oxide. How much oxygen was left over after the reaction was
complete?

From convervation of mass in chemical processes, all the reactants combined
have the same mass as all the products.
$${m}_{react}={m}_{prod}$$
$${m}_{Hg}+{m}_{{O}_{2}}={m}_{HgO}+{m}_{{O}_{2}}$$
$$114.0g+12.8g=123.1g+{m}_{{O}_{2}}$$
Solving for *m _{O2}* = 3.70 g

** Example:** 2.76 Benzene, ethane, and ethylene are just three of
a large number of hydrocarbons, compounds containing only carbon and hydrogen.
Show how the following data are consistent with the law of multiple proportions.

Compound | Mass or Carbon in 5.00 g sample | Mass of hydrogen in 5.00 g sample |
---|---|---|

Benzene | 4.61 g | 0.39 |

Ethane | 4.00 g | 1.00 |

Ethylene | 4.29 g | 0.71 |

The law of multiple proportion states that compounds are composed of
integral multiple of simple proportions. For example, if we have a
compound which has one carbon and one hydrogen, the mass ratio between
C and h are 12:1. If one carbon and two hydrogen, then 12:2 or 6:1.
If one carbon and three hydrogen, then 12:3 or 4:1. These are due to
the fact that the molar mass of carbon is 12 and of hydrogen is 1.
From these, benzene the ratio is 4.61:0.39 which reduces to be
11.8:1. Given the error in the experiment, we can say it is 12:1. It
means that benzene has 12:1 mass ratio for C to H. Indeed benzene's
chemical formula is C_{6}H_{6} therefore 6(12):6(1) or
12:1.

Ethane has 4:1, and its chemical formula is C_{2}H_{6}.
The simplest numerical ration is CH_{3}.
Using the chemical formula C_{2}H_{6} 2(12):6(1) = 24:6 = 4:1,
which is the same as the CH_{3}.

Similar analysis can be done on ethylene, C:H = 4.29g:0.71g = 6:1,
which is one carbon and two hydrogens. Ethylene's chemical formula is
C_{2}H_{4}, therefore reduces to CH_{2} as the
simplest numerial ratio.

** Example:** 2.100 How many protons, neutrons, and electrons are in
each of the following atoms?
a. ${}_{7}{}^{15}N$
b. ${}_{27}{}^{60}\mathrm{Co}$
c. ${}_{53}{}^{131}I$
d. ${}_{58}{}^{148}\mathrm{Ce}$

a. The subscript gives the number of protons, *Z*, the nuclear charge.
An atom is electrically neutral, meaning that the numbers of protons and
electrons are the same.
The superscript is the mass number, *A*, which is given by
*A = Z + N* where *N* is the number of neutrons.
From these, the number of protons = 7, number of neutrons = 8, and the number
of electrons = 7.

b. Using the arguments in a. we have *Z* = 27, *N* = 33,
# e^{-} = 27

c. *Z* = 53, *N* = 78, # e^{-} = 53

d. *Z* = 58, *N* = 90, # e^{-} = 58

** Example:** 2.121 Which of the following bonds are likely to be
covalent and/or ionic? Explain. a. B—Br b. Na—Br
c. Br—Cl d. O—Br

Determination of whether the chemical bond is covalent is nature or ionic in nature is simply to look at the periodic table. If the two atoms in question appear far apart on the periodic table, it is likely to have ionic bond. If they appear nearby, it is likely to be covalent.

a. From this assessment, B—Br is ionic.

b. Na—Br is ionic.

c. Br—Cl is covalent.

d. O—Br is covalent.

** Example:** 2.124 How many protons and electrons are in
each of the following ions?
a. $B{e}^{2+}$
b. $R{b}^{1+}$
c. $S{e}^{2-}$
d. $A{u}^{3+}$

a. Atomic number of Be is 4. So, it means that Be atom has four electrons.
In Be^{2+} ion, there are only 2 electrons. The superscript, 2+,
indicates that the ion is missing 2 electrons.

b. *Z* = 37, # of e^{-} = 36

c. *Z* = 34, # of e^{-} = 36

d. *Z* = 79, # of e^{-} = 76

** Example:** 2.140 What are the formulas of the compounds
formed from the following ions?

a. Ca^{2+} and Br^{-}
b. Ca^{2+} and SO_{4}^{2-}
c. Al^{3+} and SO_{4}^{2-}

When compounds are electrically neutral. So, when combine the two ions, you must make the compound comes out to be electrically neutral by thinking of integral multiples of ions.

a.Since Ca^{2+} has **+2** charge, you need a **-2** charge
to cancel out. Since Br^{-} has **-1** charge, you need two
Br^{-} ions. Therefore,
the chemical formula is CaBr_{2}.

b. CaSO_{4}

c. Al_{2}(SO_{4})_{3}