1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Serhud [2]
3 years ago
13

Calculate the empirical formula for each natural flavor based on its elemental mass percent composition.

Chemistry
1 answer:
algol [13]3 years ago
7 0
The complete question is;
Calculate the empirical formula for each of the following naturalflavors based on their elemental mass percent composition.

Q1)
methyl butyrate (component of apple taste andsmell):  C -58.80 %  H- 9.87 % 
O -31.33.%Express your answer as a chemical formula.


Q2)
vanillin (responsible for the taste and smellof vanilla):  C - 63.15%  H- 5.30 % 
O - 31.55%Express your answer as a chemical formula.

Q1)
empirical formula is the simplest ratio of whole number of elements in the compound. as the percentages have been given, lets calculate for 100 g of compound 
                                          C                         H                         O
mass                             58.80 g                  9.87 g                   31.33
molar mass                   12 g/mol                 1 g/mol                 16 g/mol
number of moles           58.80/12                9.87/1                    31.33/16
                                      = 4.9                      =9.87                     = 1.95
then divide number of moles by least number of moles - 1.95 in this case
                                      4.9/1.95 = 2.51      9.87/1.95 = 5.06    1.95/1.95= 1
next multiply by 2 to get numbers that can be rounded off to whole numbers
                                       2.51x2 = 5.02        5.06x2 = 10.12      1x2 = 2
when rounded off to the nearest whole number 
C - 5
 H - 10
 O - 2
therefore empirical formula is C₅H₁₀O₂

Q2) for this too since elemental composition has been given in percentages lets calculate for 100 g of compound 
                                          C                         H                         O
mass                               63.15 g                5.30 g                 31.55 g
molar mass                     12 g/mol              1 g/mol                16 g/mol
number of moles             63.15/12             5.30/1                  31.55/16
                                        =5.26                  =5.30                   =1.97
divide the number of moles by the least number of moles - 1.97
                                        5.26/1.97            5.30/1.97               1.97/1.97 
                                        =2.67                   = 2.69                      = 1
multiply each by 3 to get numbers that can be rounded off to whole numbers
                                        2.67x3 = 8.01     2.69x3 = 8.07         1x3 = 3
rounded off to the nearest whole numbers 
C - 8
H - 8
O - 3
empirical formula = C₈H₈O₃
You might be interested in
Q 9 The table lists the steps to clean water for drinking purpose. 1. Adding chlorine tablets to the water.2. Pouring the water
love history [14]

Answer:

1 and 2

Explanation:

when we r adding chlorine to water

the water is clean

8 0
3 years ago
How many atoms are found in 3.45g of CO2?
USPshnik [31]

<u>Answer:</u> The number of carbon and oxygen atoms in the given amount of carbon dioxide is 4.72\times 10^{22} and 9.44\times 10^{22} respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of carbon dioxide gas = 3.45 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in above equation, we get:

\text{Moles of carbon dioxide gas}=\frac{3.45g}{44g/mol}=0.0784mol

1 mole of carbon dioxide gas contains 1 mole of carbon and 2 moles of oxygen atoms.

According to mole concept:

1 mole of a compound contains 6.022\time 10^{23} number of molecules

So, 0.0784 moles of carbon dioxide gas will contain 1\times 0.0784\times 6.022\times 10^{23}=4.72\times 10^{22} number of carbon atoms and 2\times 0.0784\times 6.022\times 10^{23}=9.44\times 10^{22} number of oxygen atoms

Hence, the number of carbon and oxygen atoms in the given amount of carbon dioxide is 4.72\times 10^{22} and 9.44\times 10^{22} respectively

3 0
4 years ago
For the reversible, one-step reaction A+B&lt;----&gt;C+D, the forward rate constant is 52.4 /mol*h and the rate constant for the
Fantom [35]

Answer:

—-—1———————-s

7 0
4 years ago
The part of an atom that is responsible for the bonding of atoms is
Cerrena [4.2K]

Answer:

the answer is electrons

Explanation:

electrons are responsible for the chemical bonding.

3 0
3 years ago
A nomenclatura para a estrutura seguinte, de acordo com o sistema da IUPAC é: (1 Ponto) CH3-CH2-CH2-CH2-CH2-CH-C-CH2-CH3 3,4-dim
REY [17]

Answer:

6 7

Explanation:

#2 Thanks hope thus help

6 0
3 years ago
Other questions:
  • How to classify hemicetal and acetal
    15·1 answer
  • how much heat, in terms in q, would it take to produce the change in temperature indicated in the picture? what is your reasonin
    8·1 answer
  • 2
    8·1 answer
  • which city would most likely be the epicenter of an earthquake : san francisco, california, st. paul, minnesota,
    11·1 answer
  • Acid rain can be destructive to both the natural environment and human-made structures. The equation below shows a reaction that
    7·2 answers
  • An opened carbonated drink is an example of what type of solution?
    7·2 answers
  • What may be expected when K &lt; 1.0? Choose the THREE correct statements. The concentration of one or more of the reactants is
    14·1 answer
  • Which elements, when they have to, can have more than
    10·1 answer
  • Explain why, if soil on a vehicle is in a clump, care<br> should be taken to keep the clump intact.
    15·1 answer
  • 2. Which is not true of an electric current?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!