I need help calculating the things that I have italicized! This is basic gen che

Question

I need help calculating the things that I have italicized! This is basic gen chem knowledge and I've forgotten how to do it. Please help!!!

A. Results for the formation of triphenylmethyl bromide

Initial mass of triphenylmethanol (in g) = 0.208 g

Initial amount of triphenylmethanol (in mol) = 0.00079 mol

Initial volume of 33% HBr solution (in mL) = 0.6 mL

Initial mass of 33% HBr solution (in g; d = 1.354 g/mL) = 0.81 g

Initial mass of HBr (in g) = 0.27 g

Initial amount of HBr (in mol) = 0.0033 mol

Limiting reagent?????

Theoretical yield of triphenylmethyl bromide (in mol)??????

Final mass of triphenyl bromide (in grams) = 0.077 g

Final amount of triphenyl bromide (in mol) = 0.00024 mol

% yield ??????

Results (color observed) of Beilstein test in part A = Blue green

Show all calculations.

Initial amount of triphenylmethanol in moles: 0.208 g / 260.3 g/mol = 0.00079 mol

Initial mass of 33% HBr solution = 0.6 mL HBr solution x 1.354 g/mL = 0.81 g

Initial mass of HBr = 0.81 g x .33 = 0.27 g HBr

Initial amount of HBr (in mol) = 0.27 g / 80.91 g/mol = 0.0033 mol

Final amount triphenyl bromide in mol = 0.077 g / 323.23 g/mol = 0.00024 mol

B. Results for the formation of triphenylmethyl methyl ether

Initial mass of triphenylmethanol (in g) = 0.203 g

Initial amount of triphenylmethanol (in mol) = 0.00078 mol

Initial volume of methanol (in mL) = 1 mL

Initial mass of methanol (in g) = 0.792 g

Initial amount of methanol (in mol) = 0.025 mol

Limiting reagent???????

Theoretical yield of triphenylmethyl methyl ether (in mol)?????????

Final mass of triphenyl methyl ether (in g) = 0.092 g

Final amount of triphenyl methyl ether (in mol) = 0.00033 mol

% yield?????

melting point (°C) = 87.4 deg C – 88.2 deg C

Show all calculations.

Initial amount of triphenylmethanol in mol = 0.203 g / 260.33 g/mol = 0.00078 mol

Initial mass of methanol = 1 mL x 0.792 g/mL = 0.792 g

Initial moles of methanol= 0.792 g / 32.04 g/mol = 0.025 mol

Final amount of triphenyl methyl ether in mol = 0.092 g / 274.36 g/mol = 0.00033 mol

Explanation / Answer

                       triphenylmethanol + methanol    -----> triphenyl methyl ether

Molar mass(g/mol)           260.333            32.04                        274.36

Number of moles              0.00078            0.025                        0.00033

According to the balanced equation,

1 mole of triphenylmethanol reacts with 1 mole of methanol  

0.00078 mole of triphenylmethanol reacts with 0.00078 mole of methanol  

So 0.025 - 0.00078 = 0.0242 moles of methanol left unreacted so which is the excess reactant.

Since all the mass of triphenylmethanol completly reacted it is the limiting reactant.

Again from the balanced equation ,

1 mole of triphenylmethanol produces 1 mole of triphenyl methyl ether

0.00078 mole of triphenylmethanol produces 0.00078 mole of triphenyl methyl ether

So mass of triphenyl methyl ether produced = number of moles x molar mass

                                                                = 0.00078 molx 274.36 (g/mol)

                                                                = 0.214 g

Therefore the theoretical yiled is 0.214 g

Given actual yield is 0.092 g

So Percentage yield = ( actual yield/ theoretical yield) x100

                              = ( 0.092 / 0.214) x100

                             = 42.99

                             ~ 43.0 %

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