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Understanding Atomic Masses, Moles, and Empirical Formulas, Study notes of Chemistry

The concepts of atomic mass, moles, and empirical formulas in chemistry. It covers the relationship between atomic mass and avogadro's number, the calculation of molar mass, and the difference between empirical and molecular formulas. The document also includes examples and answers for determining the empirical and molecular formulas of various compounds.

What you will learn

  • How is the average atomic mass of an element calculated?
  • What is the difference between empirical and molecular formulas?
  • How can you determine the molar mass of a compound?

Typology: Study notes

2021/2022

Uploaded on 09/12/2022

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Chapter 6: Standard Review Worksheet
1. 1 amu = 1.66 _ 10–24 g. For example, the average atomic mass of sodium is 22.99 amu,
which represents the average mass of all the sodium atoms in the world (including all the
various isotopes and their relative abundances). So that we will be able to use the mass of a
sample of sodium to count the number of atoms of sodium present in the sample, we
consider that every sodium atom in a sample has exactly the same mass (the average
atomic mass). The average atomic mass of an element is typically not a whole number of
amu’s because of the presence of the different isotopes of the element, each with its own
relative abundance. Since the relative abundance of an element can be any number, when
the weighted average atomic mass of the element is calculated, the average is unlikely to be
a whole number.
2. On a microscopic basis, one mole of a substance represents Avogadro’s number
(6.022 _ 1023) of individual units (atoms or molecules) of the substance. On a macroscopic
basis, one mole of a substance represents the amount of substance present when the molar
mass of the substance in grams is taken (for example, 12.01 g of carbon will be one mole of
carbon).
3. The molar mass of a compound is the mass in grams of one mole of the compound (6.022 _
1023 molecules of the compound) and is calculated by summing the average atomic masses
of all the atoms present in a molecule of the compound. For example, a molecule of the
compound H3PO4 contains three hydrogen atoms, one phosphorus atom, and four oxygen
atoms. The molar mass is obtained by adding up the average atomic masses of these atoms:
Molar mass H3PO4 = 3(1.008 g) + 1(30.97 g) + 4(16.00 g) = 97.99 g.
4. When a new compound is prepared (the formula is not known), the percent composition must
be determined on an experimental basis. An elemental analysis must be done of a sample of
the new compound to see what mass of each element is present in the sample. For example, if
a 1.000-g sample of a hydrocarbon is analyzed, and it is found that the sample contains
0.7487 g of C, then the percent by mass of carbon present in the compound is
0.7487 g
C
1.000-g sam
p
100% = 74.87% C
Since we can use the formula of a known compound to calculate the percent composition
by mass of the compound, it is not surprising that we can go in the opposite
direction—from experimentally determined percent compositions for an unknown
compound, we can calculate the formula of the compound.
Answer Key
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Chapter 6: Standard Review Worksheet

  1. 1 amu = 1.66 _ 10–24^ g. For example, the average atomic mass of sodium is 22.99 amu, which represents the average mass of all the sodium atoms in the world (including all the various isotopes and their relative abundances). So that we will be able to use the mass of a sample of sodium to count the number of atoms of sodium present in the sample, we consider that every sodium atom in a sample has exactly the same mass (the average atomic mass). The average atomic mass of an element is typically not a whole number of amu’s because of the presence of the different isotopes of the element, each with its own relative abundance. Since the relative abundance of an element can be any number, when the weighted average atomic mass of the element is calculated, the average is unlikely to be a whole number.
  2. On a microscopic basis, one mole of a substance represents Avogadro’s number (6.022 _ 10 23 ) of individual units (atoms or molecules) of the substance. On a macroscopic basis, one mole of a substance represents the amount of substance present when the molar mass of the substance in grams is taken (for example, 12.01 g of carbon will be one mole of carbon).
  3. The molar mass of a compound is the mass in grams of one mole of the compound (6.022 _ 1023 molecules of the compound) and is calculated by summing the average atomic masses of all the atoms present in a molecule of the compound. For example, a molecule of the compound H 3 PO 4 contains three hydrogen atoms, one phosphorus atom, and four oxygen atoms. The molar mass is obtained by adding up the average atomic masses of these atoms: Molar mass H 3 PO 4 = 3(1.008 g) + 1(30.97 g) + 4(16.00 g) = 97.99 g.
  4. When a new compound is prepared (the formula is not known), the percent composition must be determined on an experimental basis. An elemental analysis must be done of a sample of the new compound to see what mass of each element is present in the sample. For example, if a 1.000-g sample of a hydrocarbon is analyzed, and it is found that the sample contains 0.7487 g of C, then the percent by mass of carbon present in the compound is 0.7487 g C 1.000-g samp

 100% = 74.87% C

Since we can use the formula of a known compound to calculate the percent composition by mass of the compound, it is not surprising that we can go in the opposite direction—from experimentally determined percent compositions for an unknown compound, we can calculate the formula of the compound.

Answer Key

  1. a. K 2 CrO 4 (molar mass = 194.2 g)

Mol K 2 CrO 4 = 5.00 g _ 1 mol 194.2 g

= 0.0257 mol K 2 CrO 4

0.0257 mol _ 6.022 10 formula units^23 1 mol

 (^) = 1.55 _ 10 22 formula units K 2 CrO 4

Atoms K = 1.55 _ 10^22 formula units _ 2 K atoms 1 formula un

= 3.10 _ 10 22 atoms K

Atoms Cr = 1.55 _ 10^22 formula units  1 Cr atom 1 formula un

= 1.55 _ 10 22 atoms Cr

Atoms O = 1.55 _ 10^22 formula units ^ 4 O atoms 1 formula un

= 6.20 _ 10 22 atoms O

b. AuCl 3 (molar mass = 303.4 g) Mol AuCl 3 = 5.00 g _ 1 mol

= 0.0165 mol AuCl (^3)

0.0165 mol AuCl 3 _ 6.022 10 formula units^23 1 mol

 (^) = 9.94 _ 10 21 formula units AuCl 3

Atoms Au = 9.94 _ 10^21 formula units _ 1 Au atom 1 formula un

= 9.94 _ 10 21 atoms Au

Atoms Cl = 9.94 _ 10 21 formula units _ 3 Cl atoms 1 formula un

= 2.98 _ 10 22 atoms Cl

c. SiH 4 (molar mass = 32.12 g) Mol SiH 4 = 5.00 g _ 1 mol

= 0.156 mol SiH 4

0.156 mol SiH 4 _ 6.022 10molecules^23 1 mol

 (^) = 9.39 _ 10 22 molecules SiH 4

Atoms Si = 9.39 _ 10 22 molecules _ 1 Si atom 1 molecul

= 9.39 _ 10 22 atoms Si

Atoms H = 9.39 _ 10^22 molecules _ 4 H atom 1 molecul

= 3.76 _ 10 23 atoms H

d. Ca 3 (PO 4 ) 2 (molar mass = 310.18 g) Mol Ca 3 (PO 4 ) 2 = 5.00 g _ 1 mol

= 0.0161 mol Ca 3 (PO 4 ) (^2)

0.0161 mol Ca 3 (PO 4 ) 2 _ 6.022 10 formula units^23 1 mol

 (^) = 9.70 _ 10 21 formula units Ca 3 (PO 4 )^2

Atoms Ca = 9.70 _ 10 21 formula units _ 3 Ca atoms 1 formula un

= 2.91 _ 10 22 atoms Ca

Atoms P = 9.70 _ 10 21 formula units _ 2 P atoms 1 formula un

= 1.94 _ 10 22 atoms P

Atoms O = 9.70 _ 10^21 formula units _ 8 O atoms 1 formula un

= 7.76 _ 10 22 atoms O