Chemistry 1 Lab Report, Lab Reports of Chemistry

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be tires since you didn’t have enough to form all the completed cars. The excess reagent is the number of reactants left after a reaction is completed which in this case would be the extra 10 car bodies. However, when the limiting reagent runs out in a chemical equation, we use something called theoretical yield to find the maximum number of products produced. Although, it’s more like a predictable amount of product based on the smallest number of moles from stoichiometric calculations. The other yields obtained from the reaction are called the actual yield and percent yield. The actual yield is the true amount for product produced by the chemical reaction. It’s always equal or less than the theoretical yield due to small amounts of product leaving the reaction or not forming any products at all. The percent yield is the actual yield over theoretical yield times 100. Solving the experiment using stoichiometry needs an example. Let’s say 5 ml of 1 M NaOH and 0.75 M CaCl 2 are mixed. The mole formula and other equations would be used: 𝑀𝑀𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 = (0.00500𝐿𝐿)(0.75 𝑚𝑚𝑀𝑀𝐶𝐶/𝐿𝐿) − 0.0038 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 𝑀𝑀𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 𝑀𝑀𝑜𝑜 𝑁𝑁𝐶𝐶𝑁𝑁𝐻𝐻 = (0.00500𝐿𝐿)(1.0 𝑚𝑚𝑀𝑀𝐶𝐶/𝐿𝐿) = 0.0050𝑚𝑚𝑀𝑀𝐶𝐶 𝑁𝑁𝐶𝐶𝑁𝑁𝐻𝐻 0.0038 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 ∗ 1 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 /1 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 = 0.0038 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 0.0050 𝑚𝑚𝑀𝑀𝐶𝐶 𝑁𝑁𝐶𝐶𝑁𝑁𝐻𝐻 ∗ 1 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 /2 𝑚𝑚𝑀𝑀𝐶𝐶 𝑁𝑁𝐶𝐶𝑁𝑁𝐻𝐻 = 0.0025 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 The limiting reactant would be NaOH due to having less product. The excess reactant would be CaCl 2 for having more product. For this experiment to be solved without using stoichiometry, you need to use test tubes and some paper. Each test tube filled with specific amounts of CaCl 2 and NaOH has water filled filter paper so they can react. The reaction gives off precipitate that you save by putting the papers in an oven. By testing each filtrate with CaCl 2 and NaOH. If the test tube goes cloudy after adding more of the compounds, then it has a limiting reagent. If nothing happens, then it has an excess reagent.

Calculations and Discussion:

1. Number of moles for NaOH (only one calculation since all liters are equal to each other): 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 = 2.50 𝑀𝑀 ∗ 0.004𝐿𝐿 = 0.01 𝑚𝑚𝑀𝑀𝐶𝐶 Number of moles for CaCl 2 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 = 0.002 𝐿𝐿 ∗ 1.0 𝑀𝑀 = 0.002 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 = 0.0035 𝐿𝐿 ∗ 1.0 𝑀𝑀 = 0.0035 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 = 0.005 𝐿𝐿 ∗ 1.0 𝑀𝑀 = 0.005 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 = 0.007 𝐿𝐿 ∗ 1.0 𝑀𝑀 = 0.007 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2
2. Theoretical yield for each experiment of Ca(OH) 2 : 0.002 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 ∗ 1 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 /1 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 = 0.002 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 0.0035 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 ∗ 1 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 /1 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 = 0.0035 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 0.005 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 ∗ 1 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 /1 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 = 0.005 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 0.007 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 ∗ 1 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 /1 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 = 0.007 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2

Multiply the number of moles Ca(OH) 2 has by its molar mass to get theoretical yield: 𝑇𝑇ℎ𝑀𝑀𝑀𝑀𝑒𝑒𝑀𝑀𝑒𝑒𝑒𝑒𝑒𝑒𝐶𝐶𝐶𝐶 𝑦𝑦𝑒𝑒𝑀𝑀𝐶𝐶𝑦𝑦 = (0.002 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 ) �74.08 (^) 𝑚𝑚𝑔𝑔𝑀𝑀𝐶𝐶� = 0.148𝑔𝑔 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 𝑇𝑇ℎ𝑀𝑀𝑀𝑀𝑒𝑒𝑀𝑀𝑒𝑒𝑒𝑒𝑒𝑒𝐶𝐶𝐶𝐶 𝑦𝑦𝑒𝑒𝑀𝑀𝐶𝐶𝑦𝑦 = (0.0035 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 ) �74.08 (^) 𝑚𝑚𝑔𝑔𝑀𝑀𝐶𝐶� = 0.259𝑔𝑔 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 𝑇𝑇ℎ𝑀𝑀𝑀𝑀𝑒𝑒𝑀𝑀𝑒𝑒𝑒𝑒𝑒𝑒𝐶𝐶𝐶𝐶 𝑦𝑦𝑒𝑒𝑀𝑀𝐶𝐶𝑦𝑦 = (0.005 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 )^ �74.08 (^) 𝑚𝑚𝑔𝑔𝑀𝑀𝐶𝐶� = 0.370𝑔𝑔 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 𝑇𝑇ℎ𝑀𝑀𝑀𝑀𝑒𝑒𝑀𝑀𝑒𝑒𝑒𝑒𝑒𝑒𝐶𝐶𝐶𝐶 𝑦𝑦𝑒𝑒𝑀𝑀𝐶𝐶𝑦𝑦 = (0.007 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 ) �74.08 (^) 𝑚𝑚𝑔𝑔𝑀𝑀𝐶𝐶� = 0.518𝑔𝑔 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 Theoretical yield for each experiment of NaOH:

0.01 𝑚𝑚𝑀𝑀𝐶𝐶 𝑁𝑁𝐶𝐶𝑁𝑁𝐻𝐻 ∗ 1 2 𝑚𝑚 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝐶𝐶^ 𝐶𝐶𝐶𝐶 𝑁𝑁(𝐶𝐶𝑁𝑁𝐻𝐻𝑁𝑁𝐻𝐻) 2 = 0.005 𝑚𝑚𝑀𝑀𝐶𝐶 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2 𝑇𝑇ℎ𝑀𝑀𝑀𝑀𝑒𝑒𝑀𝑀𝑒𝑒𝑒𝑒𝑒𝑒𝐶𝐶𝐶𝐶 𝑦𝑦𝑒𝑒𝑀𝑀𝐶𝐶𝑦𝑦 = (0.005 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀) �74.08 (^) 𝑚𝑚𝑔𝑔𝑀𝑀𝐶𝐶� = 0.370𝑔𝑔 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶(𝑁𝑁𝐻𝐻) 2

Test #3: 𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 𝑀𝑀𝑜𝑜 𝑁𝑁𝐶𝐶𝑁𝑁𝐻𝐻 𝐶𝐶𝑀𝑀𝑜𝑜𝑒𝑒 → 0.005 ∗ 2 𝑚𝑚𝑀𝑀𝐶𝐶 𝑁𝑁𝐶𝐶𝑁𝑁𝐻𝐻/1 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 = 0.01 𝑚𝑚𝑀𝑀𝐶𝐶 𝐸𝐸𝐸𝐸𝑒𝑒𝑀𝑀𝑀𝑀𝑀𝑀 𝑒𝑒𝑀𝑀𝐶𝐶𝑔𝑔𝑀𝑀𝑟𝑟𝑒𝑒 = 𝐼𝐼𝑟𝑟𝑒𝑒𝑒𝑒𝑒𝑒𝐶𝐶𝐶𝐶 𝐴𝐴𝑚𝑚𝑀𝑀𝐴𝐴𝑟𝑟𝑒𝑒 𝑀𝑀𝑜𝑜 𝑁𝑁𝐶𝐶𝑁𝑁𝐻𝐻 − 𝐴𝐴𝑚𝑚𝑀𝑀𝐴𝐴𝑟𝑟𝑒𝑒 𝐶𝐶𝑀𝑀𝑟𝑟𝑀𝑀𝐴𝐴𝑚𝑚𝑀𝑀𝑦𝑦 𝑀𝑀𝑜𝑜 𝑁𝑁𝐶𝐶𝑁𝑁ℎ 𝐸𝐸𝐸𝐸𝑒𝑒𝑀𝑀𝑀𝑀𝑀𝑀 𝑒𝑒𝑀𝑀𝐶𝐶𝑔𝑔𝑀𝑀𝑟𝑟𝑒𝑒 = 0.01 𝑚𝑚𝑀𝑀𝐶𝐶 − 0.01 𝑚𝑚𝑀𝑀𝐶𝐶 = 0 𝑚𝑚𝑀𝑀𝐶𝐶 𝐼𝐼𝑟𝑟 𝑔𝑔𝑒𝑒𝐶𝐶𝑚𝑚𝑀𝑀: 0 𝑚𝑚𝑀𝑀𝐶𝐶 ∗ 𝑚𝑚𝑀𝑀𝐶𝐶𝐶𝐶𝑒𝑒 𝑚𝑚𝐶𝐶𝑀𝑀𝑀𝑀 𝑀𝑀𝑜𝑜 𝑁𝑁𝐶𝐶𝑁𝑁𝐻𝐻 � (^80) 𝑚𝑚𝑔𝑔𝑀𝑀𝐶𝐶� = 0𝑔𝑔 Test #4: (The limiting reagent for this test switches from NaOH to CaCl 2 so instead of finding the amount of NaOH, we find the amount of CaCl 2 ):

𝑚𝑚𝑀𝑀𝐶𝐶𝑀𝑀𝑀𝑀 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 𝐶𝐶𝑀𝑀𝑜𝑜𝑒𝑒 → 0.01 ∗ 1 𝑚𝑚𝑀𝑀𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 /2 𝑚𝑚𝑀𝑀𝐶𝐶 𝑁𝑁𝐶𝐶𝑁𝑁𝐻𝐻 = 0.005𝑚𝑚𝑀𝑀𝐶𝐶 𝐸𝐸𝐸𝐸𝑒𝑒𝑀𝑀𝑀𝑀𝑀𝑀 𝑒𝑒𝑀𝑀𝐶𝐶𝑔𝑔𝑀𝑀𝑟𝑟𝑒𝑒 = 𝐼𝐼𝑟𝑟𝑒𝑒𝑒𝑒𝑒𝑒𝐶𝐶𝐶𝐶 𝐴𝐴𝑚𝑚𝑀𝑀𝐴𝐴𝑟𝑟𝑒𝑒 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 − 𝐴𝐴𝑚𝑚𝑀𝑀𝐴𝐴𝑟𝑟𝑒𝑒 𝐶𝐶𝑀𝑀𝑟𝑟𝑀𝑀𝐴𝐴𝑚𝑚𝑀𝑀𝑦𝑦 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 𝐸𝐸𝐸𝐸𝑒𝑒𝑀𝑀𝑀𝑀𝑀𝑀 𝑒𝑒𝑀𝑀𝐶𝐶𝑔𝑔𝑀𝑀𝑟𝑟𝑒𝑒 = 0.007 𝑚𝑚𝑀𝑀𝐶𝐶 − 0.005 𝑚𝑚𝑀𝑀𝐶𝐶 = 0.002 𝑚𝑚𝑀𝑀𝐶𝐶 𝐼𝐼𝑟𝑟 𝑔𝑔𝑒𝑒𝐶𝐶𝑚𝑚𝑀𝑀: 0.002 𝑚𝑚𝑀𝑀𝐶𝐶 ∗ 𝑚𝑚𝑀𝑀𝐶𝐶𝐶𝐶𝑒𝑒 𝑚𝑚𝐶𝐶𝑀𝑀𝑀𝑀 𝑀𝑀𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 2 � (^110) 𝑚𝑚𝑔𝑔𝑀𝑀𝐶𝐶� = 0.22𝑔𝑔

5. Theoretical Yield for each reaction mixture based on # 2: Test #1: Theoretical yield: 0.002 mol Test #2: Theoretical yield: 0.0035 mol Test #3: Theoretical yield: 0.005 mol Test #4 (Theoretical yield same as test #3 because limiting reagent switched from CaCl NaOH): 2 to

Theoretical yield: 0.005 mol

The limiting reagent in the experiment would be NaOH. The calculations based on the experiment prove it due CaCl 2 produces more moles that NaOH at the last test. In the experiment, the amount of CaCl 2 was the same for each test tube. Due to putting the same amounts of NaOH and CaCl 2 in each test tube, the mass or theoretical yield of the product grew more and more. It started to plateau in the beginning because the theoretical yield using CaCl 2 was produces less product, making it the limiting reagent. As the mass increases, the theoretical yield using CaCl 2 surpassed the theoretical yield using NaOH switching the limiting reagent to NaOH.

7. Percent yields for mixtures: Percent yield for mixture %yield=Actual yield / Theoretical yield * 100 Test Tube #1 0.35g / 0.148g * 100 = 236.5% Test Tube #2 0.37g/0.259g100=142.8% Test Tube #3 0.6g/0.370g100=162.1% Test Tube #4 0.83g/0.370g*100=224.3%

0

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Mass of Ca(OH)

2 (g)

Volume of CaCl 2 (mL)

mass of Ca(OH)