Question

Équivalent MW(8/mol) Density(g/cm2) Mass (g or ml) 6.6 g mmol Réactants 50 1.02 130.14 Ethyl Acetoacetate 0.38g 1.93 g 12.5 0.88 30.03 paraformaldehyde 25 1.17 77.08 Ammonium acetate 25 ml 18 water Result table Theoretical Experimental Percentage Theoretical MW yield (mmol) yield (mg) yield ( mg ou 253.29 Product

please complete both tables and show your calculations

thanks

mass of the product 2.419 g

Answer 1

ANSWER:

The general reaction is

H, H. Eto _H:0 Ft NH40Ac

The complete balanced reaction is

， ་ ལ་བ , ༡༠ དང་། མ ་ ཡ ད༌ H20 ། + NH40ACHAH + NH Y OEt + + 3 0 H,0 OH CH1003 130.14 g/mo CH+ONCHz0 7 7.08 g/mo| 30.03g/mol C13HདON 253.59 g/mdl

Now, we can complete the first table

• The moles are calculated with the equation:

$mmoles=\frac{mass\, (g)}{MW\, (g/mol)}\times \frac{1000\, mmol}{1\, mol}$

• The equivalent for each reactant is obtained from the balanced reaction

Reactants	MW (g/mol)	Density (g/cm3)	Mass (g or mL)	mmol	Equivalent
Ethyl acetoacetate	130.14 g/mol	1.02 g/mL	6.6 g	50.71 mmol	$\frac{1\, mmol\, product}{2\, mmol\, EA}\Rightarrow 50.71\, mmol\, EA\times \frac{1\, mmol\, product}{2\, mmol\, EA}=25.34 \, mmol\, product$
Paraformaldehyde	30.03 g/mol	0.88 g/mL	0.38 g	12.65 mmol	$\frac{1\, mmol\, product}{1\, mmol\, PF}\Rightarrow 12.65\, mmol\, EA\times \frac{1\, mmol\, product}{1\, mmol\, PF}=12.65 \, mmol\, product$
Ammonium acetate	77.08 g/mol	1.17 g/mL	1.93 g	25.04 mmol	$\frac{1\, mmol\, product}{1\, mmol\, AA}\Rightarrow 25.04\, mmol\, EA\times \frac{1\, mmol\, product}{1\, mmol\, AA}=25.04 \, mmol\, product$
Water	18 g/mol	1 g/mL	25 mL	--	---

From previous table, we know that limiting reactant is the paraformaldehyde. This reactant produces only 12.65 mmol of product. This is also the theoretical yield os reaction:

	MW (g/mol)	Theoretical yield (mmol)	Theoretical yield (mg or g)	Experimental yield mg or g)	Percentage
Product	253.29 g/mol	12.65 mmol	3.204 g	2.419 g	75.50 %

• Theoretical yield in mg or g:

$Theoretical\, yield\,(g) =mmol\, product \times \frac{1\, mol}{1000\, mmol}\times MW\, (g/mol)$

$Theoretical\, yield\,(g) =12.65\, mmol \times \frac{1\, mol}{1000\, mmol}\times 253.29\frac{g}{mol}$

$\mathbf{Theoretical\, yield\,(g) =3.204\, g}$

• Percentage:

$Percentage =\frac{Exp.\, yield}{Theo.\, yield}\times 100\%$

$Percentage =\frac{2.419\, g}{3.204\, g}\times 100\%=\mathbf{75.50\, \%}$