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Bad White [126]
3 years ago
9

Isobutyl propionate is the substance that provides the flavor for rum extract. combustion of a 1.152 g sample of this carbon-hyd

rogen-oxygen compound yields 2.726 g co2 and 1.116 g h2o. what is the empirical formula of isobutyl propionate?
Chemistry
1 answer:
Ulleksa [173]3 years ago
4 0

Answer;

C7H14O2

Solution;

Isobutyl contains , oxygen, carbon and hydrogen (total mass is 1.152 g)

Mass of carbon = 12/44 × 2.726 g

                           = 0.743455 g

Mass of Hydrogen  = 2/18 × 1.116 g

                            =  0.124 g

Mass of oxygen = 1.152 - (0.7435 + 0.124)

                           =  0.2845 g

Moles of carbon ;  0.7435/12 = 0.06196 moles

Moles of hydrogen; 0.124/1 = 0.124 moles

Moles of oxygen;  0.2845/16 = 0.01778 moles

Ratios ; 0.06196/0.01778 ; 0.124/0.01778 : 0.01778/0.01778

         =  3.5 :  7.0 : 1

To make them whole numbers ; we multiply the ratios by 2  to get;

(3.5 :  7.0 : 1 )2 = 7 : 14 : 2

Thus, the empirical formula of Isobutyl propionate  is C7H14O2


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Answer:

Your strategy here will be to use the molar mass of potassium bromide,

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So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.

Potassium bromide is an ionic compound that is made up of potassium cations,

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For Br:

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=

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+

79.904 g mol

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≈

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−

So, if one mole of potassium bromide has a mas of

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m it follows that three moles will have a mass of

3

moles KBr

⋅

molar mass of KBr



119 g

1

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=

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You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs

mass of 3 moles of KBr

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

360 g

a

a

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∣

−−−−−−−−−

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