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Crank
3 years ago
12

Complete combustion of 7.40 g of a hydrocarbon produced 22.4 g of CO2 and 11.5 g of H2O. What is the empirical formula for the h

ydrocarbon?
Show work please?
Chemistry
1 answer:
cluponka [151]3 years ago
3 0
<span>C2H5 First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2. Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass. moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule. Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen. moles C = 0.50899 moles H = 0.638361 * 2 = 1.276722 We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon. total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185 7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked. Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen. 0.50899 / 1.276722 = 0.398669 0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5. Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is C2H5</span>
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Which are the hybrid orbitals of the carbon atoms in the following molecules? (a) H3C―CH3 sp sp2 sp3 (b) H3C―CH═CH2 sp sp2 sp3 (
kherson [118]

Answer:

(a)  sp³    sp³

   H₃<u>C</u> - <u>C</u>H₃

(b)     sp³           sp²

     H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂

                sp²

(c)     sp³        sp    

    H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH

              sp         sp³

(d)     sp³    sp²    

       H₃<u>C</u> - <u>C</u>H=O

Explanation:

Alkanes or the carbons with all the single bonds are sp³ hybridized.

Alkenes or the carbons with double bond(s) are sp² hybridized.

Alkynes or the carbons with triple bond are sp hybridized.

Considering:

(a) H₃C-CH₃ , Both the carbons are bonded by single bond so both the carbons are sp³ hybridized.

Hence,

 sp³    sp³

H₃<u>C</u> - <u>C</u>H₃

(b) H₃C-CH=CH₂ , The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp² hybridized because they are bonded by double bond.

Hence,

   sp³           sp²

H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂

         sp²

(c) H₃C-C≡C-CH₂OH , The carbons of the methyl group and alcoholic group are sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp hybridized because they are bonded by triple bond.

Hence,

   sp³        sp    

H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH

         sp         sp³

(d)CH₃CH=O, The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The other carbon is sp² hybridized because it is  bonded by double bond to oxygen.

Hence,

   sp³    sp²    

 H₃<u>C</u> - <u>C</u>H=O

     

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