### Avogadro Number Calculations II How Many Atoms or Molecules?Problems 11 - 25

Problem #11: A solution of ammonia and water contains 2.10 x 1025 water molecules and 8.10 x 1024 ammonia molecules. How many total hydrogen atoms are in this solution?

Solution:

Ammonia's formula is NH3 and water's is H2O. 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 1024) = 2.43 x 1025 H atoms

2) Water's contribution:

(2) (2.10 x 1025) = 4.20 x 1025 H atoms

3) Sum them up:

2.43 x 1025 + 4.20 x 1025 = 6.63 x 1025 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:

Al2(SO4)3 18H2O

666.4134 g/mol

2) Convert 4.080 g of Al2(SO4)3 18H2O to moles:

4.080 g / 666.4134 g/mol = 0.006122326 mol

3) Determine formula units of Al2(SO4)3 18H2O in the 4.080 g:

(0.006122326 mol) (6.022 x 1023 formula units / mol) = 3.68686 x 1021 formula units

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

(3.68686 x 1021 formula units) (18 water molecules per formula unit) = 6.636 x 1022 water molecules (answer to a)

5) In 3.68686 x 1021 formula units of Al2(SO4)3 18H2O, there are a total of 30 oxygen atoms in each formula unit:

(3.68686 x 1021 formula units) (30 O atoms per formula unit) = 1.106 x 1023 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 = NH4Cl

Mass of one mole of ammonium chloride = 53.4916 g

Mass of 4 moles of H = 4.0316 g

Fraction H in NH4Cl = 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 NH4Cl.

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 1023 mole¯1) = 9.00 x 1023 atoms

Solution #2:

1) Convert grams to moles:

20.0 g / 53.4916 g/mol = 0.3738905 mol

2) Convert moles to number of NH4Cl formula units:

(0.3738905 mol) (6.022 x 1023 formula units / mole) = 2.2515686 x 1023 formula units of NH4Cl

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

(2.2515686 x 1023 formula units) (4 atoms / form. unit) = 9.01 x 1023 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/cm3. Answer in units of atoms.

Solution:

1) Determine mass of mercury in 9.70 cm3:

(9.70 cm3) (13.55 g/cm3) = 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 1023 atoms/mol) = 3.94 x 1023 atoms

Problem #15: What is the mass of CH4 molecules that can be made from 15.055 x 1023 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 1023

x = 3.011 x 1023 atoms of C

3) Since there is 1 atom of C for every 1 molecule of CH4, we have:

3.011 x 1023 molecules of CH4

4) Calculate moles of CH4:

3.011 x 1023 molecules / 6.022 x 1023 molecules/mol = 0.5000 mole of CH4

5) Calculate mass:

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

Problem #16: 8.0213 g of CH4 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 CH4 are present:

(0.5000 mol) (6.022 x 1023 molecules/mol) = 3.011 x 1023 molecules

3) Determine total atoms:

(3.011 x 1023 molecules) (5 atoms/molecule) = 15.055 x 1023 atoms

Based on the fact that CH4 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 kPa

Note 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 H2 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 1023 molecules/mol) = 7.2912 x 1022 molecules

4) Determine number of atoms:

(2 atoms/molecule) (7.2912 x 1022 molecules) = 1.46 x 1023 atoms

Problem #18: How may protons are there in six moles of NaNO3?

Solution:

1) Determine number of protons in one formula unit of NaNO3:

Na ---> 11
N ---> 7
three O ---> 24

total ---> 42

2) Determine protons in one mole of NaNO3:

(42 protons / formula unit) (6.022 x 1023 formula unit / mol) = 2.52924 x 1025 protons / mol

3) Determine protons in six moles of NaNO3:

(6 mol) (2.52924 x 1025 protons / mol) = 1.517544 x 1026 protons

1.518 x 1026 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 C14H18N2O5, 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 1023 molecules/mol) = 1.43232 x 1020 molecules

3) Determine number of hydrogen atoms in 1.43232 x 1020 molecules of aspartame:

(1.43232 x 1020 molecules) (18 atoms/molecule) = 2.58 x 1021 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 ---> 5

There would be no other changes to the solution steps.

Solution using dimensional analysis:

 1 mol 6.022 x 1023 molecules 18 atoms of H 0.0700 g x ––––––––– x ––––––––––––––––––– x –––––––––––– = 2.58 x 1021 atoms 294.3052 g 1 mol 1 molecule

Problem #20: What mass of carbon is present in 4.806 x 1026 molecules of C2H5OH?

Solution:

1) Determine moles of ethyl alcohol present:

4.806 x 1026 molecules / 6.022 x 1023 molecules/mol = 798.07373 mol

2) There are 2 moles of C in every mole of C2H5OH:

798.07373 mol x 2 = 1596.1475 mol of C

3) Determine mass of C:

(1596.1475 mol) (12.011 g/mol) = 19171.3276225 g

19.17 kg

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

 4.806 x 1026 molecules EtOH 1 mol EtOH 2 mol C 12.011 g C 1 kg ––––––––––––––––––––––– x ––––––––––––––––––––––– x –––––––––– x –––––––– x ––––– = 19.17 kg C 1 6.022 x 1023 molecules EtOH 1 mol EtOH 1 mol C 1000 g

Problem #21: In exactly 1 mole of the hydrate CuSO4 5H2O, how many grams are present of (a) the hydrate, (b) the anhydrate, and (c) water.

Solution:

1) In one mole of the hydrate, there is present this:

one mole of CuSO4
five moles H2O

2) Determine the molar mass for the hydrate. Remember to include the five waters.

249.681 g

This is the answer to (a)

3) Determine the molar mass of the anhydrate:

159.607 g

This is the answer to (b)

Remember, the anhydrate does not have any water in it, hence the formula is CuSO4.

4) Determine the total mass of water in one mole of the hydrate:

(18.015 g/mol) (5 mol) = 90.074 g

This is the answer to (c).

You could have also done this:

249.681 g − 159.607 g = 90.074 g

Problem #22: How many grams of carbon are in 30.0 g of CH3COOH?

Solution using dimensional analysis:

A common abbreviation for CH3COOH 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 CH3COOH weighs 60.05 g

one mole of CH3COOH contains 2 moles of C, which weigh 24.022 g

the 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 CH3COOH?

Solution using dimensional analysis:

A common abbreviation for CH3COOH 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 C5H4; 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 = 2

2 times C5H4 = C10H8

64.0866 is the mass of one mole of the empirical formula of C5H4.

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 C10H8) = 0.11548065 mol C

4) Determine number of atoms of C:

(0.11548065 mol C) (6.022 x 1023 atoms C /mol C) = 6.954 x 1022 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 C10H8, 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 1024. Determine the mass of chromium in this sample.

Solution:

1) The formula for potassium dichromate is K2Cr2O7. 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 1024 atoms / 11 atoms/form. unit = 1.936364 x 1023 formula units

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

(1.936364 x 1023 formula units) (2 atoms Cr / form. unit) = 3.872728 x 1023 atoms Cr

4) Determine moles of Cr:

3.872728 x 1023 atoms / 6.022 x 1023 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/cm3, 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 cm3

2) Determine mass of calcium carbonate in sample:

(207.4747 cm3) (2.711 g/cm3) = 562.4639 g

3) The formula for calcium carbonate is CaCO3. 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 1023 form. units/mol = 3.38422 x 1024

5) There are three O atoms per formula unit of CaCO3. Determine the number of oxygen atoms:

(3.38422 x 1024 formula units) (3 atoms / form. unit) = 1.015 x 1025 atoms

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

 207.4747 cm3 2.711 g 1 mol 6.022 x 1023 formula units 3 atoms ––––––––––– x –––––– x ––––––––– x ––––––––––––––––––––– x –––––––––––– = 1.015 x 1025 atoms 1 1 cm3 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 (C6H14) if a sample contains 1.00 x 1024 total atoms? The density of hexane is 0.6548 g/cm3.

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 1024 atoms

x = 5.00 x 1022 molecules

2) Determine moles of hexane:

5.00 x 1022 molecules / 6.022 x 1023 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/cm3 = 10.9 cm3 (to three sig figs)