All categories

3 years ago

The empirical formula of an organic compound is C2H4O. The molecular mass of the compound is 176g/mol.

Chemistry

1 answer:

Brrunno [24]3 years ago

6 0

Answer:

The molecular formula of the compound is $C_{8}H_{16}O_{4}$ . The molecular formula is obtained by the following expression shown below

$\textrm{Molecular formula }= n\times \textrm{Empirical formula}$

Explanation:

Given molecular mass of the compound is 176 g/mol

Given empirical formula is $C_{2}H_{4}O$

Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.

Empirical formula mass of the compound = $\left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}$

$n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4$

$\textrm{Molecular formula }= n\times \textrm{Empirical formula}$

Molecular formula = 4 $\times$ $C_{2}H_{4}O$

Molecular formula is $C_{8}H_{16}O_{4}$

You might be interested in

A student performed a chemical reaction in which 35 grams of hydrogen and 65 grams of oxygen reacted to form water. What is the

sergejj [24]

Answer:

The mass of the product is 73,06 g

Explanation:

The reaction to form water is:

2H2 + O2 ---> 2H2O

35 g 65 g

Molar mass H2 = 2g/m ---> I have 17,5 moles (35g / 2 g/m)

Molar mass O2 = 32g/m ---> I have 2,03 moles (65g / 32g/m)

2 moles of H2 __reacts with __ 1 mol O2

17,5 moles of H2 __ reacts with ___ (17,5 m . 1 m) / 2 m = 8,75 m

I have just 2,03 m of O2 and I need 8,75 m so O2 is my limiting reagent

1 mol of O2 ___ reacts with 2 moles H2

2,03 moles of O2 __ reacts with (2,03 m . 2 m) / 1 m = 4,06 m

I have 17,5 moles of H2 and I need 4,06 m. H2 is my excess reagent

<u><em>"All operations are done with the limiting reagent"</em></u>

1 mol O2 are required____ to form 2 H2O

2,03 mol O2 are required __ to form (2,03m . 2m) / 1m = 4,06 m

Molar mass of water: 18 g/m

Mass of water ----> Molar mass . Moles water = 18 g/m . 4,06 m = 73,08g

8 0

3 years ago

A 151.5-g sample of a metal at 75.0°C is added to 151.5 g at 15.1°C. The temperature of the water rises to 18.7°C. Calculate the

Kryger [21]

Answer:

The specific heat capacity of the metal is 0.268 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 151.5 grams

The temperature of the metal = 75.0 °C

Temperature of water = 15.1 °C

The temperature of the water rises to 18.7°C.

The specific heat capacity of water is 4.18 J/°C*g

Step 2: Calculate the specific heat capacity of the metal

heat lost = heat gained

Q = m*c*ΔT

Qmetal = - Qwater

m(metal) * c(metal) * ΔT(metal) = m(water) * c(water) * ΔT(water)

⇒ mass of the metal = 151.5 grams

⇒ c(metal) = TO BE DETERMINED

⇒ΔT( metal) = T2 - T1 = 18.7 °C - 75.0 °C = -56.3 °C

⇒ mass of the water = 151.5 grams

⇒ c(water) = 4.184 J/g°C

⇒ ΔT(water) = 18.7° - 15.1 = 3.6 °C

151.5g * c(metal) * -56.3°C = 151.5g * 4.184 J/g°C * 3.6 °C

c(metal) = 0.268 J/g°C

The specific heat capacity of the metal is 0.268 J/g°C

5 0

3 years ago

A____is a prediction about the outcome of a test or experiment.

Annette [7]

Answer:

hypothesis is the answer

Explanation:

4 0

3 years ago

Read 2 more answers

Identify the product of radioactive decay and classify the given nuclear reactions accordingly.

marta [7]

Answer: a) $_{88}^{234}\textrm{Ra}\rightarrow _{86}^{230}\textrm{Ra}+_{2}^{4}\textrm{He}$

b) $_{92}^{238}\textrm{Ra}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}$

c) $_{92}^{238}\textrm{Ra}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}$

d) $_{6}^{14}\textrm{C}\rightarrow _{7}^{14}\textrm{N}+_{-1}^{0}{\beta}$

e) $_{12}^{24}\textrm{Mg}\rightarrow _{12}^{24}\textrm{Mg}+_{0}^{0}{\gamma}$

Explanation:

A balanced nuclear equation is one in which the atomic number and mass number remains same on both sides of the equation i.e the number of protons and neutrons remain same.

General representation of an element is given as:

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

a) $_{88}^{234}\textrm{Ra}\rightarrow _{86}^{230}\textrm{Ra}+_{2}^{4}\textrm{He}$

b) $_{92}^{238}\textrm{U}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}$

c) $_{94}^{242}\textrm{Pu}\rightarrow _{92}^{238}\textrm{Ra}+_{2}^{4}\textrm{He}$

d) $_{6}^{14}\textrm{C}\rightarrow _{7}^{14}\textrm{N}+_{-1}^{0}{\beta}$

e) $_{12}^{24}\textrm{Mg}\rightarrow _{12}^{24}\textrm{Mg}+_{0}^{0}{\gamma}$

3 0

3 years ago

Which of the following statements is true?

eduard

Answer:

Lines of latitude are used to divide Earth into 24 time zones

Explanation:

5 0

3 years ago