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
kolezko [41]
3 years ago
10

Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxyg

en.
In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?

For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (10 points)
Chemistry
1 answer:
inysia [295]3 years ago
4 0

Answer:

Mass of C = 47.37g

Mass of H = 10.59g

Mass of O = 42.04g

The total mass of these elements is 100g, taking a proportion of their molar masses.

C = 47.37/12= 3.95

H = 10.59/1 = 10.59

O = 42.04/16= 2.63.

Dividing through with the smallest proportion which is 2.63

C=3.95/2.63 = 1.5

H =10.59/2.63 =4

O = 2.63/2.63= 1

Multiplying through by 2 to get a whole number.

C = 1.5x2 = 3

H= 4x2 = 8

O = 1x2= 2

The empirical formula is C3H6O2

(Empirical formula)n= molecular mass

(C3H8O2)n =228.276

(12x3 +8+16x2)n= 228.276

76n = 228.276

n = 228.276/76

n = 3

Molecular formula = Empirical formula

=(C3H8O2)3 = C9H24O6

The molecular formula is C9H24O6

You might be interested in
Write the following number correct to 2 significant figures: 3644000<br><br> Hurry Plz NO TIME!!!
Nadusha1986 [10]
3644000 rounded to 2 significant figures is equivalent to 3.6
8 0
2 years ago
Read 2 more answers
A 2.1−mL volume of seawater contains about 4.0 × 10−10 g of gold. The total volume of ocean water is about 1.5 × 1021 L. Calcula
Fiesta28 [93]

Answer:

Total worth of gold in the ocean = $5,840,000,000,000,000

Explanation:

As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.

Therefore, In 1 L of ocean water there will be,

(4.0 x 10^-10)/0.0021

= 1.9045 x 10^-7 g of gold per Liter of ocean water.

So in 1.5 x 10^-21 L of ocean water, there will be

(1.9045 x 10^-7) * (1.5 x 10^-21)

= 2.857 x 10^14 g of gold in the ocean.

1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is

20.44 * (2.857 x 10^14)

= $5,840,000,000,000,000

8 0
3 years ago
Which of the following is most likely to occur if the organ labeled 1 stops working correctly?
Dafna1 [17]

theres no link or pictureeee

6 0
3 years ago
The CNO cycle produces energy in which of the following?
d1i1m1o1n [39]
Stars on the main sequence fuse hydrogen into helium via a six-stage sequence of reactions
7 0
3 years ago
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0
3 years ago
Other questions:
  • Smog usually forms from gound-level ozone and what other human-made pollutant?
    10·2 answers
  • A solution with the concentration of 0.15 M is made using 8.00 grams of K2CrO4. How many milliliters of H2O should be used?
    6·1 answer
  • What are all of the living and nonliving things in an area called?
    10·2 answers
  • You decide to boil water to cook noodles. You place the pan of water on the stove
    9·1 answer
  • You are driving a Winnebago at a casual speed of 54 miles per hour (mph) on a journey to visit your grandmother in Alabama.
    6·1 answer
  • Please help! <br> I will make you brainliest!!!
    11·1 answer
  • You start with 1 L of CO2 at standard temperature and pressure in a closed container. If you raise the temperature of the gas, t
    10·1 answer
  • Answer EIGHT questions.
    11·1 answer
  • How might the information on melting point, boiling point, and flammability be useful to materials scientists
    12·1 answer
  • Condensed structural formula of benzoic acid
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!