Transcript Chemistry Lesson 26 Empirical Formula

Lesson 26
Empirical Formula
Objectives:
-
The student will determine an empirical,
or simplest, formula from percentage
composition data.
PA Science and Technology Standards: 3.4.10.A;
3.4.12.A
PA Mathematics Standards: 2.2.11.A, 2.3.11.C,
2.4.11.E, 2.5.11.A
I.
Formulas can be determined from composition
data
a.
When a new substance is discovered, the
discoverers don’t usually know the formula
b.
They do know the percentage composition
c.
This can be used to determine the formula, by
converting everything from percentages to moles.
d.
These moles can then be divided by the smallest
number that is determined, and this will give
small, whole numbers which can be used to
determine ratios.
e.
Example
i. As part of a science fair project, Antonio is
analyzing the contents of fresh alkaline batteries. He has
determined that one ingredient is a black powdery
compound of 63% manganese and 37% oxygen. What is
the compound’s formula?
List the percentages
Mn = 63%
O = 37%
Determine the number of moles of each element present – do
this by assuming you have 100 g of the compound in question.
1 mol Mn
63 g Mn X -------------- = 1.1467 --- = 1.1 mol Mn
54.94 g Mn
37 g O
1 mol O
X ------------ = 2.3125 --- = 2.3 mol O
16 g O
Select the smallest number of moles, and divide all your number
of moles by that one:
1.1 mol Mn
--1.1
= 1 mol Mn
2.3 mol O
--1.1
= 2.1 mol O
1.0 mol Mn and 2.1 (which is close to 2) mol O translates into:
MnO2 – the formula written in smallest whole numbers.
II.
Recap:
a.
% to grams (assume 100g)
b.
grams to moles
c,
divide by the smallest
III.
Special cases
a.
Sometimes when you get to the last step, you don’t
get numbers close to whole numbers. The
following chart should help you with this:
i. 0.1-0.2 – round down.
ii. 0.3 – multiply everything by 3
iii. 0.4-0.6 – multiply everything by 2
iv. 0.7 – multiply everything by 3
v.
0.8-0.9 – round up
b.
When completed, you should have moles close
enough to whole numbers to make an empirical
formula.
IV. Examples:
Empirical Formula Calculation
The empirical formula is the formula with the lowest whole number ration between
the elements. Example: the empirical formula of glucose which is C6H12O6 would
be CH2O.
1. To calculate the empirical formula from the percent composition:
1.
Change the % signs to grams.
2.
Convert grams to moles
3.
Divide all the numbers of moles by the lowest number of moles.
4. This gives you the ratio of the number of elements in the empirical formula.
Example: 11% Hydrogen and 89% Oxygen
11g H and 89 g O
11g H ( 1 mole/ 1.0079g) = 10.91378 moles
89 g O( 1 mole/ 15.999g) = 5.562848 moles
Oxygen has the smaller number of moles
H 10.91378/ 5.562848 = 1. 96 = about 2
O 5.562848/5.562848 = 1
Ratio is 2 H’s for every 1 O so the empirical formula is H O
a.
To calculate the empirical formula from experimental
data:
a. convert the grams given into moles for each
element and follow same procedure as above.
Empirical Formula Problems
Objectives:
-
-
The student will convert percentage composition data
into an empirical formula.
The student will use supplied data to calculate a missing
piece of data to complete the problem.
PA Mathematics Standards: 2.2.11.A, 2.4.11.E
Level 1
Determine the simplest formula for each compound listed below. In the
first eight problems, the percentage composition of the compound is given.
In the last two, laboratory data for the compound are presented.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
80.0% carbon, 20.0 % hydrogen
71.5% calcium, 28.5% oxygen
82.2% nitrogen, 17.8% hydrogen
85.7% carbon, 14.3% hydrogen
6.6% aluminum, 93.4% iodine
92.2% carbon, 7.8% hydrogen
23.5% potassium, 76.5% iodine
50.0 % sulfur, 50.0% oxygen
An oxide of arsenic contains 3.26 g of arsenic and 1.04 g of
oxygen. What is the empirical formula for this oxide?
A sample of sodium oxide weighing 12.57 g contains 9.34 g of
sodium. What is the empirical formula for this compound?
Level 2
Determine the simplest formula for each compound listed below. In the first
six problems, the percentage composition of the compound is shown. In the
last four, laboratory data for the compound are given.
1.
2.
3.
4.
5.
6.
7.
8.
9.
63.5% silver, 8.2% nitrogen, 28.2% oxygen
14.3% nitrogen, 4.1% hydrogen, 81.6% bromine
24.7% potassium, 34.7% manganese, 40.5% oxygen
56.6% potassium, 8.68% carbon, 34.7% oxygen
9.92% carbon, 58.7% chlorine, 31.4% fluorine
37.8% carbon, 6.4% hydrogen, 55.8% chlorine
35.0% nitrogen, 5.0% hydrogen, 60.0% oxygen
27.4% sodium, 1.2% hydrogen, 14.3% carbon, 57.1% oxygen
Analysis of a sample of a sulfur acid shows it to contain 0.17 g of
hydrogen, 2.82 g of sulfur, and 5.67 g of oxygen. What is the
simplest formula for this compound?
10. Analysis of a salt results in the following composition: 3.47 g of
sodium, 2.12 g of nitrogen, and 7.27 g of oxygen. What is the
empirical formula for this salt?
11. A barium salt is found to contain 21.93 g of barium, 5.12 g of
sulfur, and 10.24 g of oxygen. What is the simplest formula of this
compound?
12. An ore containing zinc, carbon, and oxygen, and weighing
485.35 g is analyzed and found to contain 46.59 g of carbon and
186.37 g of oxygen. What is the simplest formula for this
compound?
Level 3
Determine the simplest formula for each compound listed below. In the first
six problems, the percentage composition of the compound is given. In the
last four, laboratory data for the compound are given.
1.
2.
3.
4.
5.
6.
7.
8.
26.6% potassium, 35.4% chromium, 38.0% oxygen
74.0% carbon, 8.7% hydrogen, 17.3% nitrogen
69.8% iron, 30.2% oxygen
83.7% carbon, 16.3% hydrogen
50.8% zinc, 16.0% phosphorus, 33.2% oxygen
21.2% nitrogen, 6.1% hydrogen, 24.3% sulfur, 48.4% oxygen
Chemical analysis of a 10.000 g sample of oil of wintergreen shows
that it consists of 6.32 g of carbon, 0.53 g of hydrogen, and 3.16 g
of oxygen. What is the simplest formula for oil of wintergreen?
An acid is analyzed in the laboratory and the following results are
obtained: 3.1% hydrogen, 31.6% phosphorus, 65.3% oxygen. What
is the simplest formula for this acid?
9.
Examination of 3.2 x 10-2 g of an unknown white powder shows
that the powder consists of an unknown amount of nitrogen, 2.6 x
10-3 g of hydrogen, 6.7 x 10-3 of phosphorus, and 1.37 x 10-2 g of
oxygen. What is the simplest formula for this compound?
10. A rock sample weighing 5.88 x 104 g is known to contain calcium,
phosphorus, and oxygen. The amount of the first two elements in
this rock is found to be 2.28 x 104 g and 1.18 x 104 g respectively.
What is the formula for the compound in this rock sample?