Question

Given 7.55 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

Answer 1

MC3H7COOC2H5: (12×6)+(1×12)+(16×2) = 116 g/mol
mC3H7COOH: (12×4)+(1×8)+(16×2) = 88 g/mol

88g....................................................116g
C3H7COOH + C2H5OH ---> C3H7COOC2H5 + H2O
7,55g..................................................X

X = (116×7,55)/88
X = 9,952g of ethyl butyrate

Answer 2

Answer:

$9.95gC_{6}H_{12}O_{2}$

Explanation:

The balanced reaction will be:

$C_{4}H_{8}O_{2}+C_{2}H_{5}OH=C_{6}H_{12}O_{2}+H_{2}O$

The molecular weight of the ethyl butyrate is:

Atomic weight C = 12$\frac{g}{mol}$

Atomic weight O = 16$\frac{g}{mol}$

Atomic weight H = 1$\frac{g}{mol}$

Molecular weight $C_{6}H_{12}O_{2}$ = (6*12)+(1*12)+(16*2)

Molecular weight $C_{6}H_{12}O_{2}$ = 116$\frac{g}{mol}$

The molecular weight of the butanoic acid is:

Atomic weight C = 12$\frac{g}{mol}$

Atomic weight O = 16$\frac{g}{mol}$

Atomic weight H = 1$\frac{g}{mol}$

Molecular weight $C_{4}H_{8}O_{2}$ = (4*12)+(1*8)+(16*2)

Molecular weight $C_{4}H_{8}O_{2}$ = 88$\frac{g}{mol}$

Finding the mass using stoichiometry:

$7.55gC_{4}H_{8}O_{2}*\frac{1molC_{4}H_{8}O_{2}}{88gC_{4}H_{8}O_{2}}*\frac{1molC_{6}H_{12}O_{2}}{1molC_{4}H_{8}O_{2}}*\frac{116gC_{6}H_{12}O_{2}}{1molC_{6}H_{12}O_{2}}=9.95gC_{6}H_{12}O_{2}$