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
Bond [772]
3 years ago
15

5.50g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 13

6./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 14.24g water 2.91g Use this information to find the molecular formula of X.
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
3 0

Answer:

The molecular formula is C8H8O2

Explanation:

Step 1: Data given

Mass of compound X = 5.50 grams

Mass of CO2 = 14.24 grams

Molar mass of CO2 = 44.01 g/mol

Mass of H2O = 2.91 grams

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Molar mass H = 1.01 g/mol

Molar mass of the compound = 136 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 14.24 grams / 44.01 g/mol

Moles CO2 = 0.324 moles

Step 3: Calculate moles C

For 1 mol cO2 we have 1 mol C

For 0.324 moles CO2 we have 0.324 moles C

Step 4: Calculate mass C

Mass C = 0.324 moles * 12.01 g/mol

Mass C = 3.89 grams

Step 5: Calculate moles H2O

Moles H2O = 2.91 grams / 18.02 g/mol

Moles H2O = 0.161 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.161 moles H2O we have 2*0.161 = 0.322 moles H

Step 7: Calculate mass H

Mass H = 0.322 moles * 1.01 g/mol

Mass H = 0.325 grams

Step 8: Calculate mass O

Mass O = 5.50 grams - 3.89 grams - 0.325 grams

MAss O = 1.285 grams

Step 9: Calculate moles O

Moles O = 1.285 grams / 16.0 g/mol

Moles O = 0.0803 moles

Step 10: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.324 moles / 0.0803 moles = 4

H: 0.322 moles / 0.0803 moles = 4

O: 0.0803 moles / 0.0803 moles = 1

The empirical formula is C4H4O

The molar mass of this empirical formula is 68 g/mol

Step 11: Calculate the molecular formula

We have to multiply the empirical formula by n

n = 136 g/mol / 68 g/mol

n = 2

We have to multiply the empirical formula by 2

Molecular formula = 2*(C4H4O) = C8H8O2

The molecular formula is C8H8O2

You might be interested in
The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.42×10−4s−1 at a certain temperature. What is t
Keith_Richards [23]

Answer:

a) Half life of the decomposition = 4951.1 s ≈ 4950 s

b) Time it will take for the concentration of SO₂Cl₂ to decrease to 25% of its initial concentration = 9900 s

c) If the initial concentration of SO₂Cl₂ is 1.00 M, time it will take for the concentration to decrease to 0.78 M is 1775s

d) If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10² s is 0.146 M

e) If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10² s is 0.1398 M

Explanation:

Let C₀ represent the initial concentration of SO₂Cl₂

And C be the concentration of SO₂Cl₂ at anytime.

a) Rate of a first order reaction is represented by

dC/dt = - KC

dC/C = - kdt

Integrating the left hand side from C₀ to C₀/2 and the right hand side from 0 to t(1/2) (where t(1/2) is the radioactive isotope's half life)

In [(C₀/2)/C₀] = - k t(1/2)

In (1/2) = - k t(1/2)

- In 2 = - k t(1/2)

t₍₁,₂₎ = (In 2)/k

t₍₁,₂₎ = (In 2)/(1.4 × 10⁻⁴)

t₍₁,₂₎ = 4951.1 s

b) dC/C = - kdt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = - kt

C/C₀ = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C = 25% of C₀ = 0.25C₀

0.25C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.25

- kt = In 0.25

- kt = - 1.386

t = 1.386/(1.4 × 10⁻⁴) = 9900 s

c) C = C₀ e⁻ᵏᵗ

C = 0.78 M; C₀ = 1.00 M

0.78 = 1 e⁻ᵏᵗ

e⁻ᵏᵗ = 0.78

- kt = In 0.78

- kt = - 0.2485

t = 0.2485/(1.4 × 10⁻⁴) = 1775 s

d) C = C₀ e⁻ᵏᵗ

C₀ = 0.150 M, t = 2 × 10² s = 200 s

C = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 200) = 0.972

C = 0.15 × 0.972 = 0.146 M

e) C = C₀ e⁻ᵏᵗ

C₀ = 0.150 M, t = 5 × 10² s = 500 s

C = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 500) = 0.9324

C = 0.15 × 0.9324 = 0.1398 M

6 0
3 years ago
Please help with this , it’s very important
timurjin [86]

Explanation:

a) Cr is <u>acid</u> K is <u>base</u>

b) Ca is <u>base</u> and the other left part is <u>acid</u>

c)Ca is <u>base</u> and F2 is <u>acid</u>

<u>d</u><u>)</u><u> </u>NH4 is <u>acid</u> and SO4 is <u>base</u>

5 0
3 years ago
In the preparation of a certain alkyl halide, 10 g of sodium bromide (NaBr), 10 mL distilled water (H20), and 9 mL 3-methyl-1-bu
Novosadov [1.4K]

Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.

The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.

Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles

We can obtain the mass of 3-methyl-1-butanol from its density.

Mass = density × volume

Density of 3-methyl-1-butanol =  0.810 g/mL

Volume of  3-methyl-1-butanol = 9 mL

Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL

Mass of 3-methyl-1-butanol = 7.29 g

Number of moles of 3-methyl-1-butanol =  mass/molar mass =  7.29 g/88 g/mol = 0.083 moles

Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol

Mass of product 1-bromo-3-methylbutane = number of moles × molar mass

Molar mass of 1-bromo-3-methylbutane = 151 g/mol

Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol

= 12.53 g

Recall that % yield = actual yield/theoretical yield × 100

Actual yield of product = 6.48 g

Theoretical yield = 12.53 g

% yield = 6.48 g/12.53 g × 100

% yield = 51.7%

Learn more: brainly.com/question/5325004

7 0
2 years ago
a piece of candy is burned in a calorimeter raising up the tempature of 500g of water from 21 C to 25 C water has a specific hea
OverLord2011 [107]
   The energy  that was  released   by the candy  is calculates using the below formula
Q=Mc delta T

Q= heat energy
m= mass (500g)
C= specific heat  capacity) = 4.18 j/g/c
 delta t =change in temperature = 25- 21 = 4 c

Q=  500 g  x 4.18 j/g/c x 4c = 8360  j

3 0
3 years ago
(a) if a sample containing 2.00 ml of nitroglycerin is detonated, how many total moles of gas are produced? (b) if each mole of
gregori [183]
Detonation of nitroglycerin: 

4C_3H_5N_3O_9  ------\ \textgreater \  12CO_2+6N_2+O_2+H_2O

<span>Mass nitroglycerin = 2.00 mL x 1.592 g/mL = 3.184 g 
</span>
Moles = mass / molar mass = 3.184<span> g/ 227.0872 g/mol = 0.01402
</span>
the ratio between nitroglycerin and Carbon dioxide is 4 : 12 
So, moles CO2 = 0.01402 x 12 / 4 =0.0420

the ratio between nitroglycerin and N2 is 4 : 6 
moles N2 = 0.01402 x 6 / 4 =0.0841

<span>the ratio between nitroglycerin and O2 is 4 : 1 </span>
moles O2 = 0.01402 x 1 / 4 = 0.0035

<span>the ratio between nitroglycerin and water is 4 : 1 </span>
<span>in the same way moles water = 0.005258 </span>

total moles = 0.0420 + 0.0841 + 0.0035 + 0.005258 = 0.130758

0.130758<span> x 55 = 5.78 L </span>

Mass N2 = 0.0841 mol x 28.0134 g/mol = 2.3548 g


6 0
3 years ago
Other questions:
  • Help me please someone?
    15·1 answer
  • How much mass is in 100 ml of water
    8·1 answer
  • What nevres carrys information from the hairs inside the cochlea?
    6·1 answer
  • What was thompson working with when he discovered the cathode rays
    8·2 answers
  • What is the activation energy of a reaction whose rate constant doubles when the temperature is increased from 300K to 350K (ass
    7·1 answer
  • Solutions that can conduct electricity are called .
    11·1 answer
  • If you are given ethanol (C8H18) 6.3lbs/ gal and you have 12 gallons how many moles do you have? 
    7·1 answer
  • Radiation is energy that is
    10·1 answer
  • Plzz help 15 points List the fossils in order from oldest to youngest in the diagram. What can you conclude about
    7·1 answer
  • Which of these is true about electrons? posses a positive electrical charge of one (+1) have a negative electrical charge of one
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!