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Grams - Moles - Number of atoms

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Regulus 77.5

on 14 December 2015

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Transcript of Grams - Moles - Number of atoms

Si
Silicon
28.09
14
Atomic number = number of protons = number of electrons
Atomic mass
28.09 amu (atomic mass units)
- the mass of one atom expressed in atomic mass units
28.09g
- the mass of one mole of an element
1 mole of Si contains atoms and weighs 28.09 grams.
6.022 x 10
23
mass
mole
# of units
We have 5g of Si.
How many moles of atoms are in the sample?
How many atoms are in 5 g of Si?
1 mole (Si) - 28.09g (Si)
? mole (Si) - 5.00g (Si)
? =
5.00g (Si) x 1 mole (Si)
28.09g (Si)
= 0.178 mole (Si)
1 mole (Si) -
6.022 x 10
23
atoms (Si)
0.178 mole (Si) - ? atoms (Si)
? =
0.178 mole (Si) x
6.022 x 10
23
atoms (Si)
1 mole (Si)
=1.07 x 10
23
atoms (Si)
or simply:
? = 0.178 mole (Si) x 6.022 x 10 atoms (Si) = 1.07 x 10 atoms (Si)
23
23
mass
mole
# of units
We have it from the periodic table.
Given to us.
1 mole of anything contains 6.022 x 10 units of that anything. In this case we are talking about atoms.
23
mass
mole
# of units
We have 3 moles of Si.
What is the mass of this sample?
How many atoms are in the sample?
1 mole (Si) - 28.09 g (Si)
3 moles (Si) - ? g (Si)
? =
3 moles (Si) x 28.09 g (Si)
1 mole (Si)
= 84.25 g (Si)
1 mole (Si) - 6.022 x 10 atoms (Si)
3 moles (Si) - ? atoms (Si)
23
? =
3 moles (Si) x 6.022 x 10 atoms (Si)
1 mole (Si)
= 18.07 x 10 atoms (Si)
23
23
mass
mole
# of units
Calculate both the number of moles in a samples of silicon containing 5.00 x 10 atoms and the mass of the sample.
20
1 mole (Si) - 6.022 x 10 atoms (Si)
? moles (Si) - 5.00 x 10 atoms (Si)
23
20
? =
5.00 x 10 atom atoms (Si) x 1 mole (Si)
6.022 x 10 atoms (Si)
20
23
= 0.83 x 10 mole (Si) = 0.00083 mole (Si)
- 3
1 mole (Si) - 28.09 g (Si)
0.00083 mole (Si) - ? g (Si)
? =
0.00083 mole (Si) x 28,09 g (Si)
1 mole (Si)
= 0.0233 g (Si)
"Mole" is a mathematical term that describes 6.022 x 10 units of something.
It is just a number. Like the word "dozen" describes 12 units of something.

If I say give me one dozen of roses - that would mean - give me 12 roses.
If I say give me one mole of roses - that would mean - give me 6.022 x 10 roses.

If I say that I have one mole of water that means that I have 6.022 x 10 molecules of water (H O) which consists of 6.022 x 10 atoms of oxygen and 2 x 6.022 x 10 = 12.044 x 10 atoms of hydrogen.

We multiply number of hydrogen atoms by 2 because there are 2 atoms of hydrogen per every atom of oxygen in water molecule (H O) and we have 6.022 x 10 molecule of water.
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23
23
23
23
23
2
2
23
So, one mole of water is made of 6.022 x 10 water molecules. Each water molecule consists of 1 atom of oxygen and 2 atoms of hydrogen. Consequently, in one mole of water, we have 6.022 x 10 oxygen atoms (1 mole of oxygen) and 12.044 x 10 hydrogen atoms (2 moles of hydrogen).
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23
23
Molar mass (MM) of a substance is the mass in grams of one mole of the substance - so - it is the sum of mass of all the atoms that it is made of.

1 mole of water is made of 1 mole of oxygen
and 2 moles of hydrogen
From periodic table we know that:

Mass of 1 mole of oxygen is 16.00g
Mass of 2 moles of hydrogen is 1.01g x 2 = 2. 02g

Molar mass (MM) of water is 16.00 + 2.02 = 18.02g/mol
Fine molar mass (MM) of:

MM (SiO ) = 28.09 + 16.00 x 2 = 60.09g/mol
MM (HCl) =
MM (H SO ) =
MM (H O ) =
MM (NaCl) =
2
4
2
2
#units = moles x 6.022 x 10
23
mass = moles x MM(molar mass)
mass = moles x MM
#units = moles x 6.022 x 10
23
2
C H O --- Juglone

1. What is the molar mass (MM) of juglone?
2. There is a sample of 1.56 x 10 g of pure juglone. How many mole of juglone does this sample represent?

MM (C H O ) =
moles (C H O ) =


How many molecules are in the sample?
#molecules =
10
6
3
10
6
3
10
6
3
CaCO --- Calcium carbonate

1. What is the molar mass (MM) of calcium carbonate?
2. A sample of calcium carbonate contains 4.86 moles. What is the mass in grams of this sample?

MM (CaCO ) =
mass (CaCO ) =
3
3
3
Significant figures (sig figs)
Counting sig figs:

1. Leading zeros never count.
2. Trailing zeros count only if there is a decimal point.
3. All numbers that aren't leading or trailing zeroes always count.


How many sig figs are in these numbers:

0.00085
8001
7.00
700
700.
6.022 x 10
83.90
0.085600
8900000000
8900.00000
5 x 10
23
8
Significant figures in calculations:

1. In addition and subtraction --- the result is reduced to the fewest decimal places.

2. In multiplication and division --- the result is reduced to the fewest number of sig figs.
23.8g + 99.98g =
98.876L - 87.7L =
4970 + 543.1 =

18.09/34 =
78.0 x 0,000028 =
6.022 x 10 x 28.1 =
23
mass
moles
# of units
mass = MM x moles
# of units = moles x Avogador's number
moles =
mass
MM
moles =
# of units
Avogadro's number
2 major ways to describe composition of a compound:
Compound formula

- describes a compound in terms of the numbers of its constituent atoms.

H O
H SO
C H OH
CO
C H O
Percent composition / mass percent

- describes a compound in terms of the percentages (by mass) of its elements.


2
2
4
2
6
12
6
Mass of each element present in 1 mole of the compound.
2
5
Mass of 1 mole of the compound - aka Molar mass (MM)
x 100%
H SO - sulfuric acid

MM (H SO ) = 1.01 x 2 + 32.07 + 16.00 x 4 = 98.09g/mol
mass of 1 mole of H SO = 98.09g

1 moles of H SO contains 2 moles of hydrogen
1 mole of sulfur
4 moles of oxygen

mass of 2 moles of hydrogen = 1.01 x 2 = 2.02g
mass of 1 mole of sulfur = 32.07g
mass of 4 moles of oxygen = 16.00 x 4 = 64g

Mass of each element present in 1 mole of the compound.
Mass of 1 mole of the compound - aka Molar mass (MM)
x 100%
mass percent of H =
mass of H in 1 mole of H SO
MM of H SO
2
4
2
4
2
4
2
4
2
4
x 100% =
2. 02g (H)
98.09g (H SO )
2
4
x 100% = 2.06%
mass percent of S =
32.07g (S)
98.09g (H SO )
x 100% = 32.69%
2
4
mass percent of O =
64.00g (O)
98.09g (H SO )
x 100% = 65.25%
2
4
2
4
Reality check: 65.25 + 32.69 + 2.06 = 100
Check yourself:

Calculate mass percent of ethanol C H OH.
Calculate mass percent of C H O.

2
5
10
14
So, why do we do all these mass percent acrobatics?

Well, we might come across an unknown compound which means we don't know its formula. Through serious of reactions, we might find out mass percent of the elements that this compound is made of.

Now the process is reverenced - we can find out formula of the compound from mass percent of its elements.

Let's say we have a compound that we know contains:
38.67% carbon (C),
16.22% hydrogen (H)
45.11% nitrogen (N).

Compound formula indicates number of atoms in the compound. We have to convert masses of elements to numbers of atoms.

Where do we get masses of elements? - in mass percent. We can choose to work with a sample of the unknown compound that weighs 100.0g

That automatically will provide us with information that in 100.0g of the compound there are:

38.67g of C,
16.22g of H
45. 11g of N

How many atoms of C in 38.67g of C?
How many atoms of H in 16.22g of H?
How many atoms of N in 45.11g of N?

mass = MM x mole
mole =
mass
MM
mole (C) =
38.67g (C)
12.01g/mole (C)
= 3.22 mol (C)
mole (H) =
16.22g (H)
1.01g/mole (o)
= 16.09 mol (O)
mole (N) =
45.11g (N)
14.01g/mole (N)
= 3.22 mol (N)
Now we can find smallest whole number ratio:
C:
3.22
3.22
= 1
H:
16.09
3.22
= 4.997 = 5
N:
3.22
3.22
= 1
So, the formula might be CH N.
But it also can be C H N or C H N . Each of them has the correct relative number of atoms.

CH N - is empirical formula of the compound - the simplest whole number ratio of atoms present in a compound.

We must know molar mass (MM) of the compound to find molecular formula.

5
2
10
2
3
15
3
5
Let's say we know that MM (unknown) = 31.06g/mol
How does it help us?

Check yourself:

Find the empirical and molecular formula for a compound that gives following percentages on analysis (in mass percents):

Cl - 71.65%
C - 24.27%
H - 4.07%

MM (unknown compound) = 98.96g/mol

Check yourself:

An unknown compound contains:
43.64% phosphorus (P) by mass.
56.36% oxygen (O) by mass.
MM (unknown compound) = 283.88g/mol.

What are the compound's empirical and molecular formulas?

Chemical equations
- all atoms present in the reactants must be accounted for among the product.

CH + O CO + H O
4
2
2
2
H + O H O
2
2
2
Empirical and molecular formulas (review):
2 s.f.
4 s.f.
3 s.f.
1 s.f.
3 s.f.
4 s.f.
4 s.f.
5 s.f.
2 s.f.
9 s.f.
1 s.f.
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