How Many Atoms or Molecules?

Problems 11 - 25

**Problem #11:** A solution of ammonia and water contains 2.10 x 10^{25} water molecules and 8.10 x 10^{24} ammonia molecules. How many total hydrogen atoms are in this solution?

**Solution:**

Ammonia's formula is NH_{3} and water's is H_{2}O. Ammonia has three atoms of H per molecule and water has two atoms of H per molecule.

1) Ammonia's contribution:

(3) (8.10 x 10^{24}) = 2.43 x 10^{25}H atoms

2) Water's contribution:

(2) (2.10 x 10^{25}) = 4.20 x 10^{25}H atoms

3) Sum them up:

2.43 x 10^{25}+ 4.20 x 10^{25}= 6.63 x 10^{25}H atoms

**Problem #12:** (a) How many water molecules are there in a 4.080 g sample of solid aluminum sulfate octadecahydrate? (b) How many oxygen atoms are there in the 4.080 g sample?

**Solution:**

1) The formula (and molar mass) of aluminum sulfate octadecahydrate is:

Al_{2}(SO_{4})_{3}⋅18H_{2}O666.4134 g/mol

2) Convert 4.080 g of Al_{2}(SO_{4})_{3} **⋅** 18H_{2}O to moles:

4.080 g / 666.4134 g/mol = 0.006122326 mol

3) Determine formula units of Al_{2}(SO_{4})_{3} **⋅** 18H_{2}O in the 4.080 g:

(0.006122326 mol) (6.022 x 10^{23}formula units / mol) = 3.68686 x 10^{21}formula units

4) Every formula unit has 18 water molecules associated with it:

(3.68686 x 10^{21}formula units) (18 water molecules per formula unit) = 6.636 x 10^{22}water molecules (answer to a)

5) In 3.68686 x 10^{21} formula units of Al_{2}(SO_{4})_{3} **⋅** 18H_{2}O, there are a total of 30 oxygen atoms in each formula unit:

(3.68686 x 10^{21}formula units) (30 O atoms per formula unit) = 1.106 x 10^{23}O atoms (answer to b)

**Problem #13:** Determine how many atoms of hydrogen are in 20.0 grams of ammonium chloride.

**Solution #1:**

1) Ammonium chloride = NH_{4}Cl

Mass of one mole of ammonium chloride = 53.4916 gMass of 4 moles of H = 4.0316 g

Fraction H in NH

_{4}Cl = 4.0316 / 53.4916 = 0.075368843 <--- the old-school name for this is 'gravimetric factor'

2) To determine the mass of hydrogen in a specific mass of ammonium chloride, multiply by the fraction of H in NH_{4}Cl.

Mass of H = (20.00 g) (0.075368843) = 1.50737686 g

3) Determine moles of hydrogen:

1.50737686 g / 1.008 g/mol = 1.49541355 mol

4) Determine atoms of hydrogen:

(1.49541355 mol) (6.022 x 10^{23}mole¯^{1}) = 9.00 x 10^{23}atoms

**Solution #2:**

1) Convert grams to moles:

20.0 g / 53.4916 g/mol = 0.3738905 mol

2) Convert moles to number of NH_{4}Cl formula units:

(0.3738905 mol) (6.022 x 10^{23}formula units / mole) = 2.2515686 x 10^{23}formula units of NH_{4}Cl

3) There are 4 atoms of hydrogen per formula unit:

(2.2515686 x 10^{23}formula units) (4 atoms / form. unit) = 9.01 x 10^{23}atoms (rounded to three sig figs)

**Problem #14:** How many atoms of mercury are present in 9.70 cubic centimeters of liquid mercury? The density of mercury is 13.55 g/cm^{3}. Answer in units of atoms.

Solution:

1) Determine mass of mercury in 9.70 cm^{3}:

(9.70 cm^{3}) (13.55 g/cm^{3}) = 131.435 g

2) Determine moles of mercury in 131.435 g:

131.435 g / 200.59 g/mol = 0.655242 mol

3) Determine atoms in 0.655242 mol:

(0.655242 mol) (6.022 x 10^{23}atoms/mol) = 3.94 x 10^{23}atoms

**Problem #15:** What is the mass of CH_{4} molecules that can be made from 15.055 x 10^{23} atoms?

**Solution:**

1) In 'x' molecules of methane there are:

'x' atoms of C

'4x' atoms of H

2) From which follows this equation:

x + 4x = 15.055 x 10^{23}x = 3.011 x 10

^{23}atoms of C

3) Since there is 1 atom of C for every 1 molecule of CH_{4}, we have:

3.011 x 10^{23}molecules of CH_{4}

4) Calculate moles of CH_{4}:

3.011 x 10^{23}molecules / 6.022 x 10^{23}molecules/mol = 0.5000 mole of CH_{4}

5) Calculate mass:

(0.500 mol) (16.0426 g/mol) = 8.021 g (to four sig figs)

**Problem #16:** 8.0213 g of CH_{4} contains how many __total__ atoms?

**Solution:**

1) Convert grams to moles:

8.0213 g / 16.0426 g/mol = 0.5000 mol

2) Determine how many molecules of CH_{4} are present:

(0.5000 mol) (6.022 x 10^{23}molecules/mol) = 3.011 x 10^{23}molecules

3) Determine total atoms:

(3.011 x 10^{23}molecules) (5 atoms/molecule) = 15.055 x 10^{23}atomsBased on the fact that CH

_{4}has five atoms per molecule.

**Problem #17:** 3.00 L of hydrogen gas at SATP would contain how many atoms of hydrogen?

**Solution:**

1) SATP stands for Standard Ambient Temperature and Pressure and has the following values:

Temperature = 25.0 °C

Pressure = 100.0 kPaNote that these values are different from STP. I found the values for SATP here. Look in the table, it's the sixth one down.

2) Use PV = nRT to determine moles of H_{2} present:

(100.0 kPa / 101.325 kPa/atm) (3.00 L) = (n) (0.08206 L atm / mol K) (298 K)n = 0.121076 mol

I converted kPa to atm because I have memorized the value for R I used. You can look up the value for R expressed in L kPa per mol K, if you wish.

3) Use Avogadro's Number to determine number of molecules:

(0.121076 mol) (6.022 x 10^{23}molecules/mol) = 7.2912 x 10^{22}molecules

4) Determine number of atoms:

(2 atoms/molecule) (7.2912 x 10^{22}molecules) = 1.46 x 10^{23}atoms

**Problem #18:** How may protons are there in six moles of NaNO_{3}?

**Solution:**

1) Determine number of protons in one formula unit of NaNO_{3}:

Na ---> 11

N ---> 7

three O ---> 24total ---> 42

2) Determine protons in one mole of NaNO_{3}:

(42 protons / formula unit) (6.022 x 10^{23}formula unit / mol) = 2.52924 x 10^{25}protons / mol

3) Determine protons in six moles of NaNO_{3}:

(6 mol) (2.52924 x 10^{25}protons / mol) = 1.517544 x 10^{26}protons1.518 x 10

^{26}protons seems like an appropriate answer. Note use of the 'rounding off with five' rule. Also, note that 6 does not dictate sig figs. It's an exact number, not a number measured experimentally.

**Problem #19:** A can of diet soft drink contains 70.0 mg of aspartame as a sweetener. Given that the formula of aspartame is C_{14}H_{18}N_{2}O_{5}, how many atoms of hydrogen are present in the 70.0 mg

**Solution using steps:**

1) Determine moles of 70.0 mg of aspartame:

0.0700 g –––––––––––––– = 0.000237848 mol 294.3052 g/mole

2) Determine number of aspartame __molecules__ in 0.000237848 mol:

(0.000237848 mol) (6.022 x 10^{23}molecules/mol) = 1.43232 x 10^{20}molecules

3) Determine number of hydrogen atoms in 1.43232 x 10^{20} molecules of aspartame:

(1.43232 x 10^{20}molecules) (18 atoms/molecule) = 2.58 x 10^{21}atoms (to three sig figs)

4) If another atom was called for, the 18 just above would be replaced by the following:

carbon ---> 14

nitrogen ---> 2

oxygen ---> 5There would be no other changes to the solution steps.

**Solution using dimensional analysis:**

1 mol 6.022 x 10 ^{23}molecules18 atoms of H 0.0700 g x ––––––––– x ––––––––––––––––––– x –––––––––––– = 2.58 x 10 ^{21}atoms294.3052 g 1 mol 1 molecule

**Problem #20:** What mass of carbon is present in 4.806 x 10^{26} molecules of C_{2}H_{5}OH?

**Solution:**

1) Determine moles of ethyl alcohol present:

4.806 x 10^{26}molecules / 6.022 x 10^{23}molecules/mol = 798.07373 mol

2) There are 2 moles of C in every mole of C_{2}H_{5}OH:

798.07373 mol x 2 = 1596.1475 mol of C

3) Determine mass of C:

(1596.1475 mol) (12.011 g/mol) = 19171.3276225 g19.17 kg

4) Dimensional analysis (EtOH is a widely-used abbreviation for ethyl alcohol):

4.806 x 10 ^{26}molecules EtOH1 mol EtOH 2 mol C 12.011 g C 1 kg ––––––––––––––––––––––– x ––––––––––––––––––––––– x –––––––––– x –––––––– x ––––– = 19.17 kg C 1 6.022 x 10 ^{23}molecules EtOH1 mol EtOH 1 mol C 1000 g

**Problem #21:** 2.552 x 10^{24} atoms of N are contained in ___(a)___ formula units of ammonium nitrate, which is also ___(b)___ moles of ammonium nitrate weighing ___(c)___ grams.

**Solution to (a):**

2.552 x 10 ^{24}atoms N1 formula unit NH _{4}NO_{3}–––––––––––––––––– x –––––––––––––––––– = 1.276 x 10 ^{24}form. units NH_{4}NO_{3}2 atoms N

**Solution to (b):**

2.552 x 10 ^{24}atoms N1 formula unit NH _{4}NO_{3}1 mol NH _{4}NO_{3}––––––––––––––––– x –––––––––––––––––– x ––––––––––––––––––––– = 2.119 mol NH _{4}NO_{3}2 atoms N 6.022 x 10 ^{23}formula units

**Solution to (c):**

2.552 x 10 ^{24}atoms N1 formula unit NH _{4}NO_{3}1 mol NH _{4}NO_{3}80.0426 g NH _{4}NO_{3}––––––––––––––––– x –––––––––––––––––– x ––––––––––––––––––––– x –––––––––––––––– = 169.6 g NH _{4}NO_{3}2 atoms N 6.022 x 10 ^{23}formula units1 mol NH _{4}NO_{3}

**Problem #22:** How many grams of carbon are in 30.0 g of CH_{3}COOH?

**Solution using dimensional analysis:**

A common abbreviation for CH_{3}COOH is HAc.

30.0 g HAc 1 mol HAc 2 mol C 12.011 g C ––––––––– x –––––––––– x –––––––– x –––––––– = 12.0 g C 1 60.05 g HAc 1 mol HAc 1 mol C

**Solution using a gravimetric factor:**

one mole of CH_{3}COOH weighs 60.05 gone mole of CH

_{3}COOH contains 2 moles of C, which weigh 24.022 gthe mass ratio of two moles of carbon to one mole of acetic acid is 24.022 / 60.05, which equals 0.40003 (this is the gravimetric factor)

the mass of carbon in 30.0 g of acetic acid is this:

(30.0 g) (0.40003) = 12.0 g

**Problem #23:** How many grams of carbon are in 60.0 g of CH_{3}COOH?

**Solution using dimensional analysis:**

A common abbreviation for CH_{3}COOH is HAc. Also, note how the only change from the previous problem is from 30.0 g to 60.0 g. Everything else (with the exception of the answer) remains unchanged.

60.0 g HAc 1 mol HAc 2 mol C 12.011 g C ––––––––– x –––––––––– x –––––––– x –––––––– = 24.0 g C 1 60.05 g HAc 1 mol HAc 1 mol C

**Problem #24:** The empirical formula of azulene is C_{5}H_{4}; the molar mass of azulene is 128.16 g/mol. How many C atoms are present in a 1.480 g sample of azulene?

**Solution:**

1) Determine the molecular formula of azulene:

128.16 g/mol / 64.0866 g/mol = 22 times C

_{5}H_{4}= C_{10}H_{8}64.0866 is the mass of one mole of the empirical formula of C

_{5}H_{4}.

2) Determine moles of azulene present:

1.480 g / 128.16 g/mol = 0.011548065 mol

3) Determine moles of carbon atoms in the 1.480 g sample:

(0.011548065 mol) (10 moles C / 1 mol C_{10}H_{8}) = 0.11548065 mol C

4) Determine number of atoms of C:

(0.11548065 mol C) (6.022 x 10^{23}atoms C /mol C) = 6.954 x 10^{22}atoms C (to four sig figs)

5) I could have reversed steps 3 and 4. This way would have used Avogadro's Number to calculate the molecules of C_{10}H_{8}, then used 10 atoms of C per molecule to get to the total number of C atoms present.

**Problem #25:** The total number of atoms in a sample of potassium dichromate is 2.130 x 10^{24}. Determine the mass of chromium in this sample.

**Solution:**

1) The formula for potassium dichromate is K_{2}Cr_{2}O_{7}. There are 11 atoms in one formula unit of potassium dichromate.

2) Determine how many formula units are present in the sample:

2.130 x 10^{24}atoms / 11 atoms/form. unit = 1.936364 x 10^{23}formula units

3) There are two atoms of Cr per formula unit. Determine the number of Cr atoms present:

(1.936364 x 10^{23}formula units) (2 atoms Cr / form. unit) = 3.872728 x 10^{23}atoms Cr

4) Determine moles of Cr:

3.872728 x 10^{23}atoms / 6.022 x 10^{23}atoms/mol = 0.643097 mol

5) Determine mass of Cr:

(0.643097 mol) (51.9961 g/mol) = 33.44 g (to four sig figs)

**Bonus Problem #1:** Consider a sample of calcium carbonate in the form of a cube measuring 59.20 mm on each edge. Given that the density of calcium carbonate is 2.711 g/cm^{3}, how many oxygen atoms does the sample contain?

**Solution:**

1) Determine the volume of the cube:

59.20 mm = 5.920 cm(5.920 cm)

^{3}= 207.4747 cm^{3}

2) Determine mass of calcium carbonate in sample:

(207.4747 cm^{3}) (2.711 g/cm^{3}) = 562.4639 g

3) The formula for calcium carbonate is CaCO_{3}. Determine the moles of calcium carbonate in the sample:

562.4639 g / 100.0869 g/mol = 5.61976 mol

4) Determine the number of formula units present:

(5.61976 mol) (6.022 x 10^{23}form. units/mol = 3.38422 x 10^{24}

5) There are three O atoms per formula unit of CaCO_{3}. Determine the number of oxygen atoms:

(3.38422 x 10^{24}formula units) (3 atoms / form. unit) = 1.015 x 10^{25}atoms

6) Dimensional analysis (starting from the volume of the cube):

207.4747 cm ^{3}2.711 g 1 mol 6.022 x 10 ^{23}formula units3 atoms ––––––––––– x –––––– x ––––––––– x ––––––––––––––––––––– x –––––––––––– = 1.015 x 10 ^{25}atoms1 1 cm ^{3}100.0869 g 1 mol 1 formula unit Showing the mm to cm conversion is easy using DA, but showing the cube to get the volume is a hassle. So, I did a preliminary calculation before using the DA.

**Bonus Problem #2:** What is the volume in mL of hexane (C_{6}H_{14}) if a sample contains 1.00 x 10^{24} total atoms? The density of hexane is 0.6548 g/cm^{3}.

**Solution:**

1) Let 'x' be the number of molecules of hexane present. Therefore:

6x = the number of C atoms

14x = the number of H atoms

2) Solve for x:

6x + 14x = 1.00 x 10^{24}atomsx = 5.00 x 10

^{22}molecules

2) Determine moles of hexane:

5.00 x 10^{22}molecules / 6.022 x 10^{23}molecules/mol = 0.0830289 mol

3) Determine grams of hexane:

(0.0830289 mol) (86.1766 g/mol) = 7.15515 g

4) Determine volume of hexane:

7.15515 g / 0.6548 g/cm^{3}= 10.9 cm^{3}(to three sig figs)