chem lab

Here is the completed lab report based on the data and questions provided in the document.

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Introduction and Pre-Lab Questions

• What kind of reaction is it when the products have more energy than the reactants? 1

  An endothermic reaction.

• When a reaction absorbs energy, what happens to the temperature of the surroundings? 2

  The temperature of the surroundings decreases.

• Write the heat of formation reaction for water. 3

  $H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(l)$

• Write the balanced chemical equation for this reaction. (Mg + HCl) 4

  $Mg(s)+2HCl(aq)\to MgC{l}_2(aq)+H_2(g)$

• Write the balanced chemical equation for this reaction. (MgO + HCl) 5

  $MgO(s)+2HCl(aq)\to MgC{l}_2(aq)+H_{2}O(l)$

• Write the heat of formation of magnesium oxide reaction. 6

  $Mg(s)+\frac{1}{2}{O}_2(g)\to MgO(s)$

• Take a look at your above reactions, how can you combine the first three to get the heat of formation of magnesium oxide reaction? 7

  You can use Hess’s Law. If you take the first reaction (Mg + HCl), reverse the second reaction (MgO + HCl), and add the heat of formation of water reaction, the intermediate products ($MgC{l}_2$, $H_2$, $HCl$, and $H_{2}O$) will cancel out, leaving you with the target heat of formation reaction for MgO.

  • Rxn 1: $Mg(s)+2HCl(aq)\to MgC{l}_2(aq)+H_2(g)$

  • Rxn 2 (Reversed): $MgC{l}_2(aq)+H_{2}O(l)\to MgO(s)+2HCl(aq)$

  • Rxn 3 (Water Formation): $H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(l)$

  • Sum: $Mg(s)+\frac{1}{2}{O}_2(g)\to MgO(s)$

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Calculations

1. Determining the heat capacity of the calorimeter.

• In what “direction” did heat flow when you mixed the hot water with the cold water? 8

  Heat flowed from the hot water into the cold water and the calorimeter.

• Write an expression to determine the amount of heat the hot water lost and calculate it. 9

  • Expression: $q_{hot}=m_{hot}\cdot c_{water}\cdot (T_{final}-T_{hot})$

  • Calculation:

    • $q_{hot}=(50\text{ g})\cdot (4.184\text{ J/(gcdotp°C)})\cdot (34.6\text{ °C}-62\text{ °C})$101010

    • $q_{hot}=(50\text{ g})\cdot (4.184\text{ J/(gcdotp°C)})\cdot (-27.4\text{ °C})$

    • $q_{hot}=-5732.08\text{ J}$

    • Heat Lost = $-q_{hot}$ = 5732.08 J

• Write an expression to determine the amount of heat the cold water gained and calculate it. 11

  • Expression: $q_{cold}=m_{cold}\cdot c_{water}\cdot (T_{final}-T_{cold})$

  • Calculation:

    • $q_{cold}=(50\text{ g})\cdot (4.184\text{ J/(gcdotp°C)})\cdot (34.6\text{ °C}-19.6\text{ °C})$121212

    • $q_{cold}=(50\text{ g})\cdot (4.184\text{ J/(gcdotp°C)})\cdot (15.0\text{ °C})$

    • $q_{cold}$ = 3138 J

• Where did “extra” heat from the hot water go? 13

  It was absorbed by the calorimeter.

• Write an expression to determine the heat capacity of the calorimeter and solve for it. 14

  • Expression:

    • $HeatLos{t}_{hot}=HeatGaine{d}_{cold}+HeatGaine{d}_{calorimeter}$

    • $-q_{hot}=q_{cold}+q_{cal}$

    • $q_{cal}=-q_{hot}-q_{cold}$

    • $C_{cal}=\frac{q_{cal}}{\Delta T_{cal}}$ (where $\Delta T_{cal}=T_{final}-T_{cold}$)

  • Calculation:

    • $q_{cal}=5732.08\text{ J}-3138\text{ J}=2594.08\text{ J}$

    • $\Delta T_{cal}=(34.6\text{ °C}-19.6\text{ °C})=15.0\text{ °C}$ 15

    • $C_{cal}=\frac{2594.08\text{ J}}{15.0\text{ °C}}$

    • $C_{cal}$ = 172.94 J/°C

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II. Determining the enthalpy change for the magnesium and hydrochloric acid reaction.

• Determine the moles of magnesium reacted. 16

  • Molar Mass of Mg = 24.305 g/mol

  • Moles Mg = $\frac{0.03\text{ g}}{24.305\text{ g/mol}}$ 17

  • Moles Mg = 0.001234 mol

• Write an expression for finding the heat released when magnesium and the hydrochloric acid reacted. Calculate the heat released. 18

  • Expression:

    • $HeatRelease{d}_{rxn}=-(HeatGaine{d}_{solution}+HeatGaine{d}_{calorimeter})$

    • $q_{rxn}=-(q_{sol}+q_{cal})$

    • $q_{rxn}=-((m_{total}\cdot c_{sol}\cdot \Delta T)+(C_{cal}\cdot \Delta T))$

  • Calculation:

    • $m_{total}=m_{HCl}+m_{Mg}=70.7\text{ g}+0.03\text{ g}=70.73\text{ g}$ 19

    • $\Delta T=T_{final}-T_{initial}=24.8\text{ °C}-20.5\text{ °C}=4.3\text{ °C}$ 20

    • $c_{sol}$ = 4.050 J/(g·°C) 21

    • $q_{sol}=(70.73\text{ g})\cdot (4.050\text{ J/(gcdotp°C)})\cdot (4.3\text{ °C})=1231.81\text{ J}$

    • $q_{cal}=(172.94\text{ J/°C})\cdot (4.3\text{ °C})=743.64\text{ J}$

    • $q_{rxn}=-(1231.81\text{ J}+743.64\text{ J})=-1975.45\text{ J}$

    • Heat Released = 1975.45 J

• Determine the enthalpy change for the reaction. 22

  • $\Delta H_1=\frac{q_{rxn}}{\text{moles Mg}}$

  • $\Delta H_1=\frac{-1975.45\text{ J}}{0.001234\text{ mol}}=-1,600,850\text{ J/mol}$

  • $\Delta H_1$ = -1600.9 kJ/mol

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III. Determine the enthalpy change for the magnesium oxide and hydrochloric acid reaction.

• Determine the moles of magnesium oxide reacted. 23

  • Molar Mass of MgO = 24.305 + 15.999 = 40.304 g/mol

  • Moles MgO = $\frac{0.37\text{ g}}{40.304\text{ g/mol}}$ 24

  • Moles MgO = 0.009180 mol

• Write an expression for finding the heat released when magnesium oxide and the hydrochloric acid reacted. Calculate the heat released. 25

  • Expression:

    • $q_{rxn}=-(q_{sol}+q_{cal})$

    • $q_{rxn}=-((m_{total}\cdot c_{sol}\cdot \Delta T)+(C_{cal}\cdot \Delta T))$

  • Calculation:

    • $m_{total}=m_{HCl}+m_{MgO}=70.7\text{ g}+0.37\text{ g}=71.07\text{ g}$ 26

    • $\Delta T=T_{final}-T_{initial}=22.5\text{ °C}-20.9\text{ °C}=1.6\text{ °C}$ 27

    • $c_{sol}$ = 4.050 J/(g·°C) 28

    • $q_{sol}=(71.07\text{ g})\cdot (4.050\text{ J/(gcdotp°C)})\cdot (1.6\text{ °C})=460.01\text{ J}$

    • $q_{cal}=(172.94\text{ J/°C})\cdot (1.6\text{ °C})=276.70\text{ J}$

    • $q_{rxn}=-(460.01\text{ J}+276.70\text{ J})=-736.71\text{ J}$

    • Heat Released = 736.71 J

• Determine the enthalpy change for the reaction. 29

  • $\Delta H_2=\frac{q_{rxn}}{\text{moles MgO}}$

  • $\Delta H_2=\frac{-736.71\text{ J}}{0.009180\text{ mol}}=-80,252\text{ J/mol}$

  • $\Delta H_2$ = -80.25 kJ/mol

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Analysis

• Use your findings above to determine the heat of formation of magnesium oxide. 30

  We will use Hess’s Law by combining the three reactions.

  1. Reaction 1 (Mg + HCl):

     $Mg(s)+2HCl(aq)\to MgC{l}_2(aq)+H_2(g)$

     $\Delta H_1=-1600.9\text{ kJ/mol}$

  2. Reaction 2 (Reversed):

     $MgC{l}_2(aq)+H_{2}O(l)\to MgO(s)+2HCl(aq)$

     $-\Delta H_2=-(-80.25\text{ kJ/mol})=+80.25\text{ kJ/mol}$

  3. Reaction 3 (Water Formation):

     $H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(l)$

     $\Delta H_3=-285.8\text{ kJ/mol}$ 31

  • Summation:

    $\Delta H_f(MgO)=\Delta H_1+(-\Delta H_2)+\Delta H_3$

    $\Delta H_f(MgO)=(-1600.9\text{ kJ/mol})+(80.25\text{ kJ/mol})+(-285.8\text{ kJ/mol})$

    $\Delta H_f(MgO)$ = -1806.45 kJ/mol

• The accepted value for the heat of formation of magnesium oxide is -601.6 kJ/mol. Find your percent error. 32

  • Expression: $PercentError=\frac{|\text{Accepted}-\text{Experimental}|}{|\text{Accepted}|}\times 100$

  • Calculation:

    • $PercentError=\frac{|(-601.6)-(-1806.45)|}{|-601.6|}\times 100$

    • $PercentError=\frac{|1204.85|}{601.6}\times 100$

    • $PercentError$ = 200.27 %

• Provide at least one possible source of your error and explain how it would have caused your error. 33

  • Source of Error: The extremely small mass of magnesium ribbon (0.03 g) used in Part II34.

  • Explanation: This tiny mass resulted in a very small temperature change (4.3 °C)35, which is difficult to measure accurately. More importantly, when calculating the molar enthalpy ($\Delta H=\frac{q}{n}$), the heat released ($-1975.45\text{ J}$) was divided by a very small number of moles (0.001234 mol). This division dramatically magnifies any small measurement error in the temperature or mass. This led to a massively inflated value for $\Delta H_1$ ($-1600.9\text{ kJ/mol}$), which was the primary contributor to the final calculated heat of formation and the 200% error. Using a larger mass (e.g., 0.3 g) would have produced a larger, more reliable temperature change and a more accurate molar enthalpy.