Using The Avogadro Number

Avogadro Number Calculations II
How Many Atoms or Molecules?
Problems 1 - 10

Avogadro's Number Practice Problems
Using Avogadro's Number And Molar Masses
Using The Avogadro Number Aleks
Using Avogadro's Number To Find Mass

Avogadro's number is the number of particles present when the amount of material is the same as the atomic weight expressed in grams.: One of the most convenient is Avogadro's number, 6.022 x 10 molecules per mole.: What it is is an impressive example of how successive dilutions can overcome a huge number, namely Avogadro's number.: Actually that's many more bits than you could register if. The value of Avogadro number (also known as Avogadro constant) is a positive natural integer as it indicates a finite and numerable quantity of elementary objects. No constant, as well as no quantity of physics, can be defined with infinite certainty, therefore with arbitrary precision. Avogadro’s number is the quantity that dictates the size of atoms and molecules. Thus, the existence of atoms was established when Perrin used Eq.(1) to ﬂnd that NA is in fact a very large number. In this experiment, we follow Perrin and examine the validity of Eq.(1) using 100 nm gold nanoparticles and 1 „m polystyrene spheres in a total of.

Problem #1: How many atoms of chlorine are in 16.50 g of iron(III) chloride?

Solution:

1) Determine moles of FeCl₃:

16.50 g / 162.204 g mol¯¹ = 0.101723755 mol

2) Determine how many formula units of iron(III) chloride are in 0.1017 mol:

0.101723755 mol x 6.022 x 10²³ = 6.1258 x 10²² formula units

3) Determine number of Cl atoms in 6.1258 x 10²² formula units of FeCl₃:

6.1258 x 10²² x 3 = 1.838 x 10²³ atoms (to 4 sig fig)

4) Set up using dimensional analysis:

1 mol	6.022 x 10²³	3 Cl atoms
16.50 g x	––––––––	x	––––––––––	x	–––––––––	= 1.838 x 10²³ chlorine atoms
162.204 g	1 mol	1 FeCl₃

Problem #2: How much does 100 million atoms of gold weigh?

Solution:

1) Determine moles of gold in 1.00 x 10⁸ (we'll assume three sig figs in the 100 million):

1.00 x 10⁸ atoms divided by 6.022 x 10²³ atoms/mol = 1.660578 x 10¯¹⁶ mol

2) Determine grams in 1.66 x 10¯¹⁶ mol of gold:

1.660578 x 10¯¹⁶ mol times 196.97 g/mol = 3.27 x 10¯¹⁴ g (to three sig fig)

Problem #3: 18.0 g of (NH₄)₂CO₃ is present. (a) How many oxygen atoms are present? (b) How many hydrogen atoms are present? (c) How many total atoms are present?

Solution:

1) Convert mass to moles:

18.0 g / 96.0852 g/mol = 0.18733374 mol

2) Determine how many formula units of ammonium carbonate are present:

(0.18733374 mol) (6.022 x 10²³ mol¯¹) = 1.128124 x 10²³ formula units

3) Determine oxygen atoms present:

(1.128124 x 10²³ formula units) (3 O atoms / formula unit) = 3.38 x 10²³ O atoms

4) Determine hydrogen atoms present:

(1.128124 x 10²³ formula units) (8 H atoms / formula unit) = 9.02 x 10²³ H atoms

5) Determine total atoms present:

(1.128124 x 10²³ formula units) (14 atoms / formula unit) = 1.58 x 10²⁴ atoms

For the total, simply count up how many atoms are in the formula unit. In this case, there are 14 total atoms in one formula unit of (NH₄)₂CO₃ (two N, eight H, one C and three O).

Problem #4: Suppose we knew that there were 8.170 x 10²⁰ atoms of O in an unknown sample of KMnO₄. How many milligrams would the unknown sample weigh?

Solution:

1) Determine how many formula units of KMnO₄ there are:

8.170 x 10²⁰ atoms divided by 4 atoms per formula unit

2.0425 x 10²⁰ formula units of KMnO₄

2) Determine moles of KMnO₄:

2.0425 x 10²⁰ formula units divided by 6.022 x 10²³ formula units/mol = 3.39173 x 10¯⁴ mol

3) Determine grams, then milligrams of KMnO₄:

3.39173 x 10¯⁴ mol times 158.032 g/mol = 0.0536 g

(0.0536 g) (1000 mg/g) = 53.6 mg

4) Dimensional analysis:

1 formula unit	1 mol	158.032	1000 mg
8.170 x 10²⁰ atoms x	––––––––––––	x	––––––––––	x	–––––––	x	–––––––	= 53.6 mg
4 atoms	6.022 x 10²³	1 mol	1 g

Problem #5: A solid sample of cesium sulfate contains 5.780 x 10²³ cesium ions. How many grams of cesium sulfate must be present?

Solution:

1) Determine how many formula units of Cs₂SO₄ must be present:

5.780 x 10²³ divided by 2 = 2.890 x 10²³

This is because there are 2 Cs atoms per one cesium sulfate formula unit.

2) Determine how many moles of Cs₂SO₄ are present:

2.890 x 10²³ divided by 6.022 x 10²³ mol¯¹= 0.479907 mol

3) Determine grams of cesium sulfate:

0.479907 mol times 361.8735 g/mol = 173.7 g

Problem #6: A sample of C₃H₈ has 4.64 x 10²⁴ H atoms. (a) How many carbon atoms does this sample contain? (b) What is the total mass of the sample?

Solution:

1) Convert from hydrogen atoms to C₃H₈ molecules:

4.64 x 10²⁴ H atoms divided by 8 H atoms per C₃H₈ molecule = 5.80 x 10²³ molecules of C₃H₈

2) C₃H₈ molecules to carbon atoms:

5.80 x 10²³ molecules times 3 C atoms per molecule = 1.74 x 10²⁴<---answer for (a)

3) Moles of C₃H₈:

5.80 x 10²³ molecules divided by 6.022 x 10²³ molecules / mole = 0.9631352 mol

4) Mass of C₃H₈:

0.9631352 mol times 44.0962 g/mol = 42.47 g

to three sig figs, this is 42.5 g <---answer for (b)

Problem #7: If a sample of disulfur hexabromide contains 6.99 x 10²³ atoms of bromine, what is the mass of the sample?

Solution:

1) We need the moles of bromine:

6.99 x 10²³ atoms divided by 6.022 x 10²³ atoms/mol = 1.160744 mol or Br atoms

2) Disulfur hexabromide's formula is S₂Br₆. That means there are 6 moles of Br for every one mole of S₂Br₆, therefore:

1.160744 mol / 6 = 0.193457 mol of S₂Br₆ present

3) The molar mass of S₂Br₆ is 543.554 g/mol:

543.554 g/mol times 0.193457 mol = 105.15 g

to three sig figs, 105 g

4) Dimensional analysis:

1 mol	1 mol	543.554 g
6.99 x 10²³ atoms x	––––––––––	x	–––––––	x	––––––––	= 105 g
6.022 x 10²³	6 mol	1 mol

Problem #8: A sample of dinitrogen trioxide contains 0.250 moles of oxygen. How many molecules of the compound are present?

Solution:

1) Calculate moles of N₂O₃:

0.250 mol O times (1 mol N₂O₃ / 3 mol O) = 0.083333 mol N₂O₃

2) Calculate molecules of N₂O₃:

0.083333 mol N₂O₃ times (6.022 x 10²³ molecules / mol) = 5.02 x 10²² molecules (to three sig figs)

Problem #9: Determine the number of oxygen atoms in 2.30 g of Al₂(SO₄)₃.

Solution:

1) Here are the steps:

(a) 2.30 g divided by molar mass of Al₂(SO₄)₃ = moles of Al₂(SO₄)₃

(b) moles of Al₂(SO₄)₃ times 6.022 x 10²³ = formula units of Al₂(SO₄)₃

2) Let's build that up in dimensional analysis style:

1 mol	6.022 x 10²³	12 O atoms
2.30 g x	–––––––––	x	––––––––––	x	–––––––	= 4.86 x 10²² oxygen atoms
342.147 g	1 mol	1
↑ step (a) ↑	↑ step (b) ↑	↑ step (c) ↑

Problem #10: How many potassium ions are there in 85.0 g of potassium sulfate (K₂SO₄)?

Solution:

1 mol	6.022 x 10²³	2 K atoms
85.0 g x	–––––––	x	––––––––––	x	–––––––––––	= 5.87 x 10²³ K atoms
174.26 g	1 mol	1 formula unit of K₂SO₄

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:

8.10 x 10²⁴ times 3 = 2.43 x 10²⁵ H atoms

2) Water's contribution:

2.10 x 10²⁵ times 2 = 4.20 x 10²⁵ H atoms

3) Sum them up:

2.43 x 10²⁵ + 4.20 x 10²⁵ = 6.63 x 10²⁵ H atoms

Avogadro's Number Practice Problems

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

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 if they are made from 15.05 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 = 15.05 x 10²³

x = 3.01 x 10²³ atoms of C

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

3.01 x 10²³ molecules of CH₄

4) Calculate moles of CH₄:

3.01 x 10²³ molecules divided by 6.02 x 10²³ molecules/mol = 0.500 mole of CH₄

5) Calculate mass:

0.500 mol times 16.0426 g/mol = 8.02 g (to three sig figs)

Problem #16: 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

Bonus Problem: A sample of HNO₃ contains twice as many atoms as there are atoms in 6.840 g of Al₂(SO₄)₃. Calculate the mass of the HNO₃ in the sample.

Solution:

Using Avogadro's Number And Molar Masses

1) We need to first determine the number of atoms in our sample of aluminum sulfate:

6.840 g / 342.147 g/mol = 0.0199914 mol

(0.0199914 mol) (6.022 x 10²³ formula units/mol) = 1.203882 x 10²² formula units

(1.203882 x ²² formula units) (17 atoms / formula unit) = 2.0466 x 10²³ atoms

2) The above calculation done in dimensional analysis style:

1 mole	6.022 x 10²³ form. units	17 atoms
6.840 g x	–––––––	x	––––––––––––––––––––	x	–––––––––	= 2.0466 x 10²³ atoms
342.147 g	1 mol	1 form. unit