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mamaluj [8]
3 years ago
13

Tartaric acid is the white, powdery substance that coats sour candies such as sour patch kids. combustion analysis of a 12.01-g

sample of tartaric acid-which contains only carbon, hydrogen, and oxygen-produced 14.08 g co2 and 4.32 g h2o. empirical formula
Chemistry
1 answer:
shusha [124]3 years ago
4 0

First we calculate the mass of carbon, hydrogen and oxygen.

Mass of Carbon;<span>
Molar mass of CO</span>₂ is 44.009g/mol  and molar mass of C = 12.011 g/mol 

<span>Mass of Carbon in 14.08g CO2</span>₂ = 12.011/44.009 x 14.08 = 3.84

<span> Mass Hydrogen;
Molar mass H</span>₂O is 18.0015g/mol and molar mass of H is 1.008and  H₂ = 2.016 

<span>Mass Hydrogen in 4.32g of H</span>₂O = 2.016/18.0015 x 4.32 = 0.48

<span> <span>Mass of  Oxygen = total mas of tartaric acid – (mass of carbon + mass of hydrogen) = 12.01 - (3.84+0.48) = 7.69</span>
Now divide carbon, hydrogen and oxygen through their respective atomic masses: </span>
<span>C = 3.84/12.011 = 0.32
<span>H = 0.48/1.008 = 0.48
<span>O = 7.69/15.999 = 0.48
Now divide by smallest: </span>
C = 0.32/0.32 = 1 </span>
H = 0.48/0.32 = 1.5 
O = 0.48/0.32 = 1.5 
multiply by 2 to get whole integers: </span>
C = 1 x 2 = 2 
H= 1.5 x 2 = 3 
<span>O= 1.5 x 2 = 3
<span>Thus, the Empirical formula of tartaric acid = C</span></span>₂H₃O₃

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2 years ago
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Explanation:

Hello!

In this case, according to the balanced chemical reaction, we can compute the grams of copper I chloride produced by each reactant, as shown below:

m_{CuCl}^{by\ Cu}=3.23molCu*\frac{2molCuCl}{2molCu}*\frac{99.0gCuCl}{1molCuCl}  =320gCuCl\\\\m_{CuCl}^{by\ Cl_2}=1.64molCl_2*\frac{2molCuCl}{1molCl_2}*\frac{99.0gCuCl}{1molCuCl}  =325gCuCl

Thus, since copper produces the fewest grams of CuCl, we infer it is the limiting reactant, therefore the correct mass of copper I chloride is 320 g.

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