PERCENT COMPOSITION OF A Experiment 4 COMPOUND IDEAS In the early nineteenth century, two French scientists argued about the composition of chemical...
View the step-by-step solution to:

Question

Question 3 in the lab (all necessary information of the lab is attached)


style="color:rgb(51,51,51);">Consider the following statements and describe the effect of each on the calculated percent magnesium in the sample (i.e. High, low, or no effect on the percent magnesium in the oxide- MgO.) in each case, give a short explanation for your reasoning.

A. All of the Mg does not burn

B. The reaction smokes for a brief period (consider what the smoke is due to)

C. all the Magnesium nitride is not converted ultimately to MgO

D. All the water is not removed from the end product.

E. Some of the product is lost by spattering.


1.0.jpeg

PERCENT COMPOSITION OF A
Experiment 4
COMPOUND
IDEAS
In the early nineteenth century, two French scientists argued about the composition of chemical
compounds. C. Berthollet asserted that the composition of a compound depended upon the proportions of
the substances reacting or the conditions of the reaction. On the other hand, J. Proust insisted that each
individual chemical compound was always made up of the same elements, which were in a definite
proportion to each other by weight. All samples of this compound, from any source, always have
exactly the same composition by weight. Let's test these hypotheses.
We are going to produce a compound, magnesium oxide, by heating magnesium in the presence of
air.
There are four aspects to this experiment: first, the chemical reactions involved in changing
magnesium to magnesium oxide by burning it in the air; second, an analysis of the class results in terms of
percent magnesium in magnesium oxide to determine the average value, most frequent value, the range
of values and the standard deviation; third, the formula of magnesium oxide based on the experimental
results; and fourth, to decide, based on the class results, whose hypothesis is supported by our results -
Berthollet's or Proust's.
Outline of Reactions
When magnesium, a very reactive metal, is burned in air, it reacts not only with oxygen to form
magnesium oxide, but also with nitrogen to form magnesium nitride.
2Mg + 02
2Mgo
3Mg + N2
Mg3 N2
Since the object of the experiment is to find out how much magnesium oxide forms from a known
weight of magnesium, it is necessary to convert all the magnesium nitride to magnesium oxide. The
following procedure brings this about.
Water is added to the mixture of magnesium oxide and magnesium nitride:
Mgo + H20 - Mg(OH) 2
get as vil borden
Mg3N2 + 6H20 -3Mg(OH)2 + 2NH3
67

1.1.jpeg

The ammonia gas is lost to the atmosphere; the product remaining in the crucible is only
magnesium hydroxide and some excess water. The final step is to heat this magnesium hydroxide to
drive off the water.
MO
Mg(OH) 2
Mgo + H20
The empirical formula can be calculated by 1) finding the weight in grams of both magnesium and
oxygen which combined; 2) calculating the number of moles of each element represented by these
weights; 3) finding the molar ratio of magnesium to oxygen; and 4) then determining the formula by
finding the smallest whole numbers which best represent this ratio.
to anna Here is a sample calculation using a different oxide:
ries Hard borelani lab
stingb s ni 92.00 g of iron reacted to form 2.86 g of iron oxide
The number of grams of oxygen is 2.86 - 2.00 = 0.86 g
gnilog oursvy
lo sorisesng ord mi rau
The number of moles of iron is 2.00 g + 56 g/mole = 0.036 moles iron
gnignore fi ba
to armed ni eflues The number of moles of oxygen is 0.86g + 16 g/mole = 0.054 moles oxygen
IsInomgqxe or no Molar ratio O/Fe = 0.054/0.036 = 1.5/1
- aflue91 quo yd balto
Smallest whole number ratio (found here by multiplying both numbers by 2) = 3/2
Formula: Fe203
Analysis
Once a set of measurements has been taken, an average of all of the results is usually presented.
Reporting the range, which is the difference between the highest and lowest values, is a means of
evaluating the precision of the measurements. A more precise picture of the value of the data comes
from a calculation of the standard deviation. This number is reported along with the average; for
example 54 15. For a reasonably large set of data, 68% of all of the values will lie within one standard
deviation of the average. Two standard deviations will include 95% of the data. Thus the magnitude of
the standard deviation shows the dispersion in the data. High precision would be indicated by a smal
standard deviation.
All scientific measurements are subject to error. These errors are grouped into two categories:
systematic errors and random errors.
Systematic errors arise from one or a combination of causes -- by a defect in the analytical
method, by an improperly functioning instrument (e.g. a meter stick improperly calibrated or a
malfunctioning balance), by impure chemicals, etc. Careful attention to detail can resolve these
problems.
68

1.10.jpeg

Computer Analysis of Mgo
1. What is the accepted value for the % Mg in Mgo?
Examine the distribution for your class and the composite curve for all classes.
your class
composite
2. a) What is the mean value?
68. 1
63.4
b) What is the most frequent value?
70
63.
c) What is the standard deviation?
2.17
2. 6
3. In what ways does the composite data differ from your class data?
The mean value of the composite data is lower than my classes
The most frequent value is higher Than The composite class (my clas.
composite- 63 ). Lastly, the standard deviation of The composite class is
higher than my class (my class - 2. 17 , composite class - 2.6)
4. Does the pooling of results in this experiment yield an average % Mg in Mgo which is
closer to the accepted value than your class average?
5. Compare the precision obtained by your class to the precision obtained by the
composite of all classes. Discuss.
High
Precision would be indicated by a small standard deviation.
The magnitude in standard deviation shows The dispersion in The data )
Thus , the precision
obtained by my class is higher because of
the lower standard deviation ) compared to The composite class
which has a lower precision relative to my class cand because it
has a hisher standard deviation)
6. What conclusion can you draw from the analysis of the data in assessing whether this
laboratory procedure gives accurate results?

1.2.jpeg

xide to
Random errors arise from unknown causes and are unavoidable. There is always some uncertainty
in every physical measurement: e.g. human judgment is involved in reading a scale or temperature
fluctuations occur during a series of measurements.
sium and
A truly random error is just as likely to be positive as negative. This fact makes the average of
ese
many measurements more reliable than any one measurement. Thus, the average % Mg in Mgo of a class
zula by
should be closer to the accepted value than the result obtained by an individual student. The average of
the results of many classes should be even closer. Again, this assumes that the errors are essentially
random and no systematic errors are present. This hypothesis can be tested by use of the normal
distribution curve (the bell curve).
If the errors are essentially random, there will be a maximum in the frequency of occurrence quite
near the accepted value for % Mg in Mgo, the curve will be symmetrical about the accepted value, and
results far from the accepted value will occur only infrequently. In other words, small errors are more
likely than large errors and positive errors are as likely to occur as negative ones.
If there are systematic errors, the average value of % Mg in Mgo will not be close to the true
value. The results will be skewed to one side of the accepted value or the other, depending on the source
of the systematic error.
Of course, sloppy technique can yield both high and low results with the average % Mg in Mgo
being close to the accepted value only by chance. But then the standard deviation will be large.
siunoasia
12
INVESTIGATION
Purpose:
To find the percent composition and empirical formula of magnesium oxide; to
evaluate experimental data.
ted.
Equipment
1 crucible and cover
1 clay triangle
1 Bunsen burner
1 glass stirring rod
es
1 ring stand
crucible tongs
1 iron ring
red litmus paper
dard
ude of
Chemicals:
magnesium metal turnings
small
Procedure:
It is important for you to follow the instructions carefully for making magnesium oxide for your
results to be meaningful.
Heat, cool, and weigh precisely a clean crucible and cover. The cover should be slightly ajar when
heating and fully closed while cooling. Weigh precisely about 0.5 g of magnesium metal into the
69

1.3.jpeg

Davidson
Institute
SCIENCE
crucible and record the weight of the crucible and magnesium with the cover on. Support the covered
crucible and contents on a clay triangle. Adjust the cover so that the crucible is slightly open to permit
some air to enter. Smoke is evidence that some product is being lost. Heat slowly. Do not allow the
reaction to smoke. Use the crucible tongs to adjust the cover so that the reaction proceeds without the
loss of material from the crucible. Continue heating until the magnesium no longer glows red, plus
another 5 minutes.
Remove the cover and heat strongly for about 5 minutes. Allow to cool. With your glass stirring
rod, carefully crush the product in the crucible into a powder. Tap any powder sticking to the stirring
rod into the crucible.
Add about 0.5 ml (10 drops) of deionized water dropwise, rinsing any specks of solid off the glass
stirring rod. Warm slowly to allow the water to evaporate without spattering. Detect the odor of
ammonia which is produced from magnesium nitride. The presence of ammonia can be confirmed by
holding moist red litmus paper over the warmed mixture and observing the color change to blue.
Partially cover the crucible and heat strongly for at least 10 minutes. Strong heating is necessary to
convert magnesium hydroxide to magnesium oxide. The bottom of the crucible should be red hot. Allow
to cool to room temperature and weigh.
Your instructor may ask you to repeat the experiment for a second value.
arly or beolb grind
Calculate the percent magnesium in your sample of magnesium oxide. Record your value on the
class data sheet, using the proper number of significant figures.
Examine the class data. Count the number of times each value for the percent magnesium appears.
On graph paper, make a bar graph using the numbers (frequency) on the y-axis against the percent
magnesium on the x-axis. This graphical picture should give an indication of the kinds of error that
were present in this experiment.

1.4.jpeg

PERCENT COMPOSITION OF A COMPOUND
DATA
Weight of crucible, cover and magnesium
19.94 9
b. Weight of crucible and cover
19. 44 9
c. Weight of magnesium (a - b)
weight of cover : 9.80 g
509
d. Weight of crucible, cover and magnesium oxide
20.19 9
e. Weight of crucible and cover (from b)
19. 4 4 9
f. Weight of magnesium oxide (d - e)
20.19-19.44
0.75 9
g.
Experimental percent magnesium in magnesium oxide
67 %
(show method of calculation clearly) Report to the proper
35 2sigfigs
number of significant figures.
Using accepted atomic weights, calculate:
h. Number of moles of magnesium atoms in (c)
0.50 "
. 02057
24, 305 1 mol
0. 021
i. Number of moles of oxygen atoms combined with the magnesium
0. 016
j. Formula of magnesium oxide from your experimental values
Mao
1, K. Theoretical % magnesium in magnesium oxide, based on the formula Mgo
60.30
use
. 1. Percent error (show method of calculation)
0 . 1055
4 5f
As soon as you complete the calculations, enter your value for % magnesium on the class data sheet.
71

1.5.jpeg

THOUGHT
1. Why is strong heating necessary at the end of the experiment?
strong heating is necessary at the end of the
experiment + so that all of the deionized
water is evaporated.
2. This question may require information of the kind that can be found in the Handbook of Chemistry
and Physics.
a. Is there more than one compound of magnesium and oxygen? If so, indicate the formulas of those
compounds containing only magnesium and oxygen.
yes-Mgo - Magnesium oxide
Mgo2 - Magnesium peroxide
b. The maximum temperature attainable in the crucible on strong heating with a bunsen burner is
approximately 800 C. Can the magnesium evaporate under the conditions of this experiment? Why or
why not?
add
move
NO, Ng will not evaporate completely under the conditions of this
experiment. The reason is because ine boiling point
of magnesium is 1090's and the melting point is 650.C. My will not
evaporate because its boiling point is too high (even though under 800"C it starts
3. Consider the following statements and describe the effect of each on the calculated percent to melt
magnesium in the sample (i.e. high, low, or no effect on the percent magnesium in the oxide). In each of
case, give a short explanation of your reasoning.
evaporate
a. All of the magnesium does not burn.
% My will decrease in the sample because part of the My did not
react. Therefore, the yelld of Mgo will be lower than expected calculated
amount (some Mg reacts with N2 to form MggN2 , the rest will beto
burned inthepresence of oxygen and form Mgo)
b. The reaction smokes for a brief period. (Consider what the smoke is due to.)
& My in Ngo will be higher Than expected- because if some smoke escapes
crucible ,the mass of Nigo measured at the end of the experiment will be
lower than what was actually produced (because the smoke escaped)
since the mass of Magnesium is recorded before the reaction, the of composition
of My in the fing!
. product will.
be higher
c. All the magnesium nitride is not converted ultimately to magnesium oxide.
Some of The My can react with No present in the air + form My3 N2
3MJ + NZ
72

1.6.jpeg

Khalsa ednov. Free Application f
Wednesd
Rachel Aronov
9/21
d. All the water is not removed from the end product.
e. Some of the product is lost by spattering.
4. a. Report the average percent magnesium using all the values of the class.
68
used
*
b. What is the range of values? highest -lowest
10
62, 66,67, 68, 69, 70, 72
72 - 62
figs
c. Calculate the standard deviation.
2. 2
5. Construct the bar graph of the class results as described in the INVESTIGATION section.
a. On the graph, mark the bars representing the most frequent value, the average value, and the
accepted value.
b. Are these three values the same when considered to 2 significant figures?
c. Is there evidence for either random or systematic error? How do you know?
There is evidence for
d. If there is evidence of systematic error, what do you suggest is the cause?

1.7.jpeg

6. a) Using your own data, is it possible to evaluate the ideas of Proust and Berthollet?
ve for al
b) Using the results of the entire class as the basis for your answer, can you decide whose hypothesis is
supported -- Berthollet's or Proust's? Explain.
7. The sample calculation for iron oxide in the IDEAS section of this experiment used known atomic
weights to calculate an empirical formula. However, early chemists did not have any references in
lass data
which they could look up atomic weights. Instead, they guessed at the formulas of compounds and
lover
measured the percent compositions of elements in compounds in order to calculate atomic weights.
no cor
Calculate an atomic weight for iron using the hypothetical formula Fe101 and the composition data
given in the example in the IDEAS section. You may assume the atomic weight of oxygen is known from
lation
other sources to be 16 amu,
posite
re:0 - 1:1
- 2 sig tops
1x 16 9 0 = 16 grams
16
48
verage
1 x X q Fe = Xgrams
X
112
Molar
of 0 3
3 x 16
48
X = 37. 33
ratio
Fe
2x 56
112#
round to 2sf
ecision
atomic weight of Fe is 37 g
8. In working the last problem, you may have noticed that a known atomic weight for oxygen was
1 St
required. Early chemists had no one to tell them the atomic weight of oxygen. They could have
in The
$ is
proposed that the atomic weight of oxygen be set equal to 1.00 amu in order to calculate an atomic
to
weight of iron relative to that of oxygen. Use the same information as before and calculate the atomic
my
D : Fe
weight of iron relative to that of oxygen using the hypothetical formula Fe101 3 six figs
0.86 9 Oxgen
atomic weigh Fe = x
3:2
reacts w/ 2 9 Iron
atomic weight oxJen = 1.00
e data
3 x 0. 86
X
x 2
3
* = 201.55)
oxygen
iron
201.55 ) = BX
3
2. 58 = 4
X= 1.03 amu
2.58 x = 4
74
X = 1.55

1.8.jpeg

Colo 111
67 1/
69111
70 HH
121
INSTRUCTOR
Chabra
3B1 DATE 9/ 23
EXPERIMENTAL VALUES
PERCENT COMPOSITION OF A COMPOUND
rounded to the whole #
Name
yoMg
42
1. Michael Lefore
66 2. 1
14.41
2. Ingrid Bauniza - Torres
70
1.9
13.61
Mean Value =
68.1 - 68
3 . Esther Zhang
4.41
4. Cheyanne Ruiz
47
1.21
5 . Ashley smith
69 0.9 10.81.
89.8
= 2. 17 4009 151
6 .
Jessica Joy
68 01 0.OL
S
n - 1
20 -1
7. Julianna Braniecki
608 0.1
10.01
8. Jabel James
70 1.9
13.601
9. Azwad Islam
69 10.9 10 .81
n = number of samples
10. Solarge Carrera
70 1.9
3.41
11. MaRiana Cabral
70
1.9 3, 4
S =
2
12. Alyssa cheraso
66
2
4.41
13. Amanda Ke Hlever
42 6.
137.21
14. Rachel Atronov
/Mg =_
68 + 2
15. Joe DeCurtis
16. Devyn Losco
689 0.1 0.01
69% 0.9 0.81
17. Mohammed Hasan
18. Isabel Flemming
70% /1.9
12.61
Number of significant figures
that should be used based
19. Tyler Mecook
72% 3.9 15 21
upon the experimental
20. Olivia Timinelli
672 1.1 121
68% OL
results
2
21.
10.01
22 .
23 .
24.
25 .
26.
27 .
* 4 = deviation from the
28.
mean
Total
1362
89.8
[(4 2)
1400

1.9.jpeg

Top Answer

Sign up to view the full answer

Why Join Course Hero?

Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.

-

Educational Resources
  • -

    Study Documents

    Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.

    Browse Documents
  • -

    Question & Answers

    Get one-on-one homework help from our expert tutors—available online 24/7. Ask your own questions or browse existing Q&A threads. Satisfaction guaranteed!

    Ask a Question