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Theoretical and
actual yields of
Name of the module: Chemistry
Assessment number: 1
Word count: 1700
Theoretical and actual yields of magnesium chloride
Aim:
The aim of the practical was to carry out the experiment by reacting magnesium with
hydrochloric acid and to calculate theoretical and actual yields of products based on
given amount of reactants.
Introduction:
One of the well-known laws in chemistry is the Law of Conservation of Mass
discovered by Russian scientist Mikhail Lomonosov in 1756 (APS News 2011) and
states that mass of the reactants is equal to the mass of products made in chemical
reactions. Therefore, the number of atoms used in the experiment will be the same as
the outcome of the reaction. This means that it is impossible to destroy or create an
atom during chemical experiment (Gorzynski-Smith 2013: 126-127).
Based on the law of conversation of mass, the reaction of magnesium with hydrochloric
acid to make magnesium chloride which is shown in the following chemical equation
as: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g), theoretically shows that there are equal
number of atoms on both sides, so the mass of atoms will stay the same.
However, what can change in the reaction are the bonds between elements, so the
formation of new bonds or degradation may occur leading to creation of different types
of molecules (Gorzynski-Smith 2016: 134-135). Moreover, for reactants to form a new
compound the collision is necessary, so they must collide energetically between each
other in order to break and form bonds, which is known as collision theory. For
example, when a water molecule is made the hydrogen atom is burned and combined
with oxygen as shown in the chemical equation: 2H2(g) + O2(g) → 2H2O(l). This shows
that reactants, in this case 2H2 and O2, are equal to the products of this reaction, which
is 2H2O, because there are 4 hydrogen and 2 oxygen atoms in both sides of the
equation as shown on figure 1. Therefore, the total mass of reactants is equal to the
mass of products as there was no gain or loss of any atom but only creation of covalent
bond to form a water molecule (McMurry, Fay and Robinson 2016: 70-71).
Figure 1. Chemical reaction between hydrogen and oxygen molecules to produce water (Online
Chemistry Book n.d.).
In addition to, stoichiometry is the field in chemistry which revolves around quantitative
relationship between reactants and products, such as the actual and theoretical
percentage yields to determine how much of products were produced during chemical
reaction (Hein and Arena 2014: 169-181). The percentage yields are quite important
area in industry and all the companies spend much time and money to improve it. This
is because no matter what is produced, they want it in large quantities and as cheap
as it is possible, so therefore they are using percentage yields in order to estimate the
amount of product they can make and the cost of the production. In addition, it is used
to make sure that the synthesized reaction at every step progress in high yields
because if at any step will be low percentage yield, it will cause unnecessary waste of
reactants and money (Gorzynski-Smith 2016: 167).
In chemistry, to demonstrate the number of atoms of each element within a compound,
the empirical and molecular formulas can be used. The empirical formula shows the
simplest and smallest whole number ratio of each element within the molecule e.g.
glucose empirical formula is CH2O (Hein and Arena 2014 133-135). In order to find out
empirical formula three things are needed: the elements, which are combined with
each other; atomic mass of each element; and either the ratio percentage or mass in
the proportion they are combined (Marry, Fay and Robinson 2016 120-121).
Example of the Percent composition of glucose:
C – 40.00% (12.011u), H – 6.7% (1.008u), O – 53.3% (15.999u) (Hein and Arena 2014:
56)
Molar mass of glucose: (4 x 12.011) + (12 x 1.008) + (6×15.999) = ??????. ??????
???????????? ???? ?????????? ???? ?? = = ??. ???? ??????????
???????????? ???? ?????????? ???? ?? = = ??. ???? ??????????
???????????? ???? ?????????? ???? 0 = = ??. ???? ??????????
???????????? ?????? ??ℎ?? ?????????????? ???? ??ℎ?? ???????????????? ????????????: = ?? = ??. ???? ≈ ?? = ??
Therefore, the empirical formula for glucose is CH2O.
At this time, the whole number ratios might be already obtained as in the example, so
the empirical formula can be written. But, if the numbers are decimals, all numbers
must be multiplied by the smallest number which will made them whole number, for
e.g. if A is 2,5 and B is 2, multiply it by 2, so the A will be equal 5 and B 4, which
represents the empirical formula as A5B4 (Marry, Fay and Robinson 2016: 120-121).
On the other hand, the molecular formula is used to represent the total number of
atoms within molecule but also arrangements of atoms for e.g. glucose molecular
formula is C6H12O6, which shows that within compound there are 6 carbons, 12
hydrogens and 6 oxygens. To determine the molecular formula, molar mass of the
compound and the empirical formula must be known, which then are applied to the
formula which looks following:
?????????? ????????
?? = = ???????????? ???? ?????????????????? ?????????????? ??????????
???????? ???? ?????????????????? ??????????????
‘n’ is used to multiply the empirical formula in order to work out molecular formula, e.g.
for empirical formula of glucose the calculated n value is 6, the equation looks
following: (CH2O)6 = C6H12O6 the molecular formula for glucose (Hein and Arena 2014:
136).
Table 1. Examples of empirical and molecular formula of different compounds.
Name of compound Empirical Formula Molecular Formula
Glucose CH2O C6H12O6
Methane CH4 CH4
Octane C4H9 C8H18
Caffeine C4H5N2O C8H10N4O2
Magnesium Chloride MgCl2 MgCl2
Ethyl Alcohol/Ethanol C2H6O C2H6O
Hydrogen peroxide HO H2O2
Sodium Phosphide Na3P Na3P
Moreover, the empirical formula for many compounds is usually identical to molecular
formula as shown in Table 1. This is because the formula cannot be simplified even
more as it already has low numbers, which cannot be divided, for e.g. MgCl2. On the
other hand, glucose molecule is made up of many atoms, so the molecular formula will
differ than empirical as the numbers can be divided, so C6H12O6 (molecular formula)
is simplified to give empirical formula of CH2O (Nivaldo 2014: 148).
Figure 2. Isomers and stereoisomers. However, in some occasion the empirical formula can be
the same for different compounds such as for glucose, fructose and galactose. This is because
these compounds are isomers and stereoisomers which means that they have identical formulas
but different structures, so therefore even their molecular formula is the same (Mason et al 2017:
39).
Empirical formula of compound with mass of 0.662 containing 0.35g of Al and O.
Al = 0.35g
O = 0.662 – 0.350 = 0.312g
0.35g
???????????? ???? ?????????? ???? ???? = = ??. ?????????? ??????????
26.982 (?????????? ???????? ???? ????)
0.312g
???????????? ???? ?????????? ???? ?? = = ??. ?????????? ??????????
15.999 (molar mass of O)
= ?? ?? ?? = ?? = ??. ?? ???? = ??
One of the answers is not a whole number, therefore it must be multiplied by 2 which
gives empirical formula of Al2O3.
Method:
At the start of the experiment, the measurements of some of the equipment and
sources were collected (shown in Table 3). Afterwards, the magnesium strip was
mixed with hydrochloric acid within the conical flask and by using syringe and the pipet;
the amount of hydrogen produced in the chemical reaction was calculated. Thirdly, the
homogenous mixture of Mg and HCl was moved to crucible, placed on the tripod and
gauze to be heated up using the Bunsen burner in order to make MgCl2 salt. Lastly,
the measurements of crucible were collected to find out how many salt was produced.
No list of reagents/equipments
Table 2. Risk assessment
Hazards Risk Description Preventive Measures
Glass equipment
The glass during experiment may
break. If it gets in contact with the
skin, it may lead to the cut and
bleeding but also may allow some
bacteria to invade the wound.
In order to prevent it, the lab coat must
be worn, to ensure most parts of the
body are covered, as well as gloves to
cover the hands and goggles to protect
eyes. Also, making sure that the glass
is not near the ledges to avoid it falling
down and break.
Bunsen burner The flame form Bunsen burner is
quite big, so therefore it may easily
get in contact with hair or hands
causing the skin or hair to burn.
To minimize the risk of getting in
contact with the flame, hair should be
tight up at all time. Furthermore, the
safe flame can be used, so it will be
clearly visible for everyone. Also, if the
Bunsen burner is not in use, it should
be turned off.
Hydrochloric
Acid
Hydrochloric Acid is corrosive which
means if it gets in contact with any
part of the body it will burn the skin
and can make damage to internal
parts of the body.
To avoid contact with hydrochloric
acid, the gloves, lab coat and goggles
must be worn at all time in order to
cover most parts of the body, therefore
minimizing risk of contact with an acid.
No reference for risk assessment
Results:
Table 3. Data collected from the experiment
Weight of Conical Flask 121.51 g
Weight of Conical Flask & Solution 135.85 g
Weight of Solution only 14.34 g
Weight of Magnesium Strip 0.02 g
Volume of gas produced 32 cm3
Conical flask after reaction 135.83 g
Weight of Crucible 28.63 g
Weight of Crucible & Salt 28.83g
Weight of Salt 0.20 g
Calculations to determine empirical formula for magnesium chloride:
Mg (molar mass) = 24.305 Cl2 (molar mass) = 2 x 35.453 = 70.906
24.305 + 70.906 = 95.211 (molar mass of MgCl2)
?????????????????????? ?????????????????????? ???? ???? = ?? 100 = ????. ????%
???????????????????? ?????????????????????? ???? ????2 = ?? 100 = ????. ????%
????. ???? ?????????? ???? ???? = = ??. ???? ??????????
????. ???? ?????????? ???? ????2 = = ??. ?? ??????????
= 1 = 2
Therefore, the empirical formula for magnesium chloride is MgCl2. Since this formula
cannot be simplified, molecular formula for MgCl2 will be the same.
Calculations to determine percentage yields of products:
No. of moles of magnesium used = = ??. ??????????−??
Balanced chemical equation for magnesium chloride:
Mg (s) + 2HCl (l) → MgCl2 (aq) + H2 (g)
No. of moles of H2 produced is equal 8.23×10-4, as the ratio between Mg and H2 is 1:1.
Percentage yield of MgCl2 and H2:
????. ???? ?????????? ???? ????????2 ???????????????? = = ??. ??????????−?? ??????????
2.10×10−3
???????????????????? ???? ????????2 produced = 8 .23×10−4 ?? 100 = ??????%
32 cm3: 1000 = 0.032dm3
0.
????. ???? ?????????? ???? ??2 ???????????????? == ??. ???????? ??????????
???????????????????? ???? ??2 ?????????????? = ?? 100 = ??????%
8.
Discussion:
The magnesium chloride is formed during single replacement reaction between Mg
(metal) and HCl (acid) which leads to the formation of an ionic bond between these
two elements (Gorzynski Smith 2016: 143-144). This is because Cl is in the seventh
group in periodic table, so it must gain only one electron to have 8 electrons in the
valence shell and become stable. On the other hand, Mg is in the second group in the
periodic table, so it must donate two electrons to become stable and achieve an octet.
Therefore, two chlorines are needed to create a compound. During redox reaction Mg
become an anion giving two electrons away (oxidation) and chlorines become cations
as they receive two electrons (reduction), one electron to each Cl element, in order to
form a magnesium chloride (MgCl2) salt (Nivaldo 2014: 250-251). Moreover, the
collision between particles is a key in chemical reaction as it allows to produce kinetic
energy, so the reactants can break and form bonds. In the Mg and HCl reaction, the
elevated temperature of the solution (exothermic reaction) and in the room speeded
up the movement of molecules, so therefore they collide with each other at increased
rate and at the same time accelerating the reaction. Moreover, more kinetic energy
was produced, so molecules were able to reach the activation energy which was
necessary for breaking bonds between HCl and formation of MgCl2 (Crowe and
Bradshaw 2014: 569-573).
According to the law of conservation of mass, when the experiment is complete, the
mass of the reactants supposed to be the same as the mass of products, so the
expected results would be 100% of each product. However, the results obtained after
experiment were much higher as 255% of MgCl2 and 162% of H2 were produced. One
of the reasons for this could be side reaction during an experiment. This is because
the experiment was done in the open space where multiple reactions occurred leaving
mixture of gasses in the air. Moreover, the conical flask was closed manually which
could have allowed some oxygen inside. Therefore, hydrogen produced during
reaction could react with oxygen forming covalent bond by sharing electrons between
them, forming water molecule (Smith 2013: 146). Furthermore, the formation of MgCl2
is exothermic reaction which means that the solution turned hot during experiment
(Lewis and Evans 2011: 216), so therefore the H2 reacting with oxygen would have
produced water in gaseous state, at the same time increasing the percentage yield of
H2 produced. In order to minimize this systematic error in future experiments, the
specialised equipment such as glovebox could be used. This is because this
equipment will allow to control pressure inside the vacuum (Yeada Group 2017), so
the atmospheric pressure can be minimized and allow to keep the solution in closed
compartment. Therefore, it will eliminate any additional reactants inside, so the
chances of hydrogen reacting with different elements will be minimal, making the
experiment more reliable and accurate.
Another reason for high percentage yield of products could be impurity. Since in the
experiment Mg was a limiting reactant (McMurry, Fay and Robinson 2016: 116-118),
a lot of HCl did not reacted.
Calculations to determine amount of HCl used:
No. of moles of magnesium used = = ??. ??????????−??
Ratio between Mg and HCl is 1:2, so: 2 x 8.23×10-4 = 1.646×10-3
1.646×10-3 x (1.008 + 35.453) = 0.06g of HCl used for 0.02g of Mg
14.34 – 0.06 = 14.28 g of HCl did not reacted
Calculations of experimental formula for MgCl2:
Mass of Cl = 0.20 (mass of salt) – 0.02 (mass of magnesium strip) = 0.18g
?????????? ???? ???? = = ??. ??????????−?? ?????????? ???? ???? = = ??. ??????????−??
8.22×10−4
?????????? ?????????????? ???? ?????? ???? = 8 .22×10−4 = ?? . ????
≈ ?? Therefore, experimental formula is MgCl6.
Furthermore, as already stated before the hydrogen could react with oxygen producing
water molecule, so it could get mixed with the HCl and dissolve an acid to produce
H3O+ + Cl- (hydronium and chloride ions) based on Arrhenius definition (Nivaldo 2014:
543). Therefore, the mixture of hydronium, chloride ions, excess of HCl and MgCl2 was
heated and most of it evaporated leaving crystal salt inside. However, because of the
high percentage yield of MgCl2, it is most likely that salt was impure which had an
impact on experimental formula of magnesium chloride. This means that human error
is another reason because product was not left for enough time to completely dry, so
the homogenous mixture of particles stated above could still be left inside, giving
incorrect results. In order to eliminate this error in the future, more time should be
allowed for the solution to fully dry. Therefore, all unwanted chemicals will have enough
time to evaporate and only desired product will remain in crucible, making the results
more accurate.
Conclusion:
The calculated percentage yields showed that when performing chemical reactions,
every little detail must be taken into consideration and every single step must be done
very accurately. This is because as looking back at the reaction of Mg and HCl a few
things, such as environment, human error or the equipment used had negative impact
on percentage yields of the product and those mistakes can be detrimental, especially
on the economy of industrial companies.
Aps News (2011) ‘This Month in Physics History’ American Physical Society [online]
11 November.
available from
[27 March 2019]
Chemistry Online Book (n.d.) Chemical bonds join atoms to one another [online]
available from
March 2019]
Crowe, J., and Bradshaw, T. (2014) Chemistry for the Biosciences. 3rd edition. New
York: Oxford University Press
Gorzynski-Smith, J. (2013) General, Organic, & Biological Chemistry. 7th edition.
New York: McGraw-Hill Education
Gorzynski-Smith, J. (2016) General, Organic, & Biological Chemistry. 3rd edition.
New York: McGraw-Hill Education
Hein, M., and Arena, S. (2014) Foundations of college chemistry. 14th edition.
Hoboken: Wiley
Lewis, R., and Evans, W. (2011) Chemistry. 4th edition. London: Palgrave Macmillan
Mason, K.A., Losos, J.B., Singer, S.R., Raven, P.H., and Johnson, G.B. (2014)
Biology. 11th edition. New York: McGraw-Hill Education
https://www.aps.org/publications/apsnews/201111/physicshistory.cfm
https://www.aps.org/publications/apsnews/201111/physicshistory.cfm
https://www.aps.org/publications/apsnews/201111/physicshistory.cfm
http://bodell.mtchs.org/OnlineBio/BIOCD/text/chapter4/concept4.3.html
http://bodell.mtchs.org/OnlineBio/BIOCD/text/chapter4/concept4.3.html
http://bodell.mtchs.org/OnlineBio/BIOCD/text/chapter4/concept4.3.html
McMurry. J.E., Fay, R.C., and Robinson J.K. (2015) Chemistry. 7th edition. Harlow:
Pearson Education
Nivaldo, T.J. (2014) Introductory Chemistry. 4th edition. London: Pearson Education
Yeada Group (2017) Yeada Group Limited Build a Partnership in Russia [online]
available from
April 2019]
http://www.yeadagroup.com.hk/en/new/new-54-519.html
http://www.yeadagroup.com.hk/en/new/new-54-519.html
http://www.yeadagroup.com.hk/en/new/new-54-519.html
http://www.yeadagroup.com.hk/en/new/new-54-519.html
http://www.yeadagroup.com.hk/en/new/new-54-519.html
http://www.yeadagroup.com.hk/en/new/new-54-519.html
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