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

Fifteen Examples

Probs 1-10

Mole Table of Contents


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: 2.552 x 1024 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 1024 atoms N   1 formula unit NH4NO3  
––––––––––––––––––  x  ––––––––––––––––––  = 1.276 x 1024 form. units NH4NO3
    2 atoms N  

Solution to (b):

2.552 x 1024 atoms N   1 formula unit NH4NO3   1 mol NH4NO3  
–––––––––––––––––  x  ––––––––––––––––––  x  –––––––––––––––––––––  = 2.119 mol NH4NO3
    2 atoms N   6.022 x 1023 formula units  

Solution to (c):

2.552 x 1024 atoms N   1 formula unit NH4NO3   1 mol NH4NO3   80.0426 g NH4NO3  
–––––––––––––––––  x  ––––––––––––––––––  x  –––––––––––––––––––––  x  ––––––––––––––––  = 169.6 g NH4NO3
    2 atoms N   6.022 x 1023 formula units   1 mol NH4NO3  

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)

Fifteen Examples

Probs 1-10

Mole Table of Contents