All categories

2 years ago

Determining the molecular formula of a hydrocarbon with a relative molecular mass of 54

Chemistry

1 answer:

frez [133]2 years ago

8 0

Answer:

C₄H₆

Explanation:

A hydrocarbon is a molecule that only contains atoms of Carbon and Hydrogen. As you remember, the molar mass of Carbon is 12g/mol and the molar mass of Hydrogen is 1g/mol.

To solve this question we can find the amount of Carbons necessary that doesn't exceed the 54g/mol. That is:

54g/mol / 12g/mol = 4.5

As the molecular formula must be given in whole numbers, the amount of Carbons is 4. This 4 carbons have a mass of:

12g/mol*4 = 48g/mol

And the hydrogens necessaries to suply the molecular mass are:

54 - 48 = 6 hydrogens

That means molecular formula is:

You might be interested in

You have 15.42g of C2H6. How many moles of H2O can be made?

Amiraneli [1.4K]

Answer: The moles of water produced are 1.54 moles.

Explanation:

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

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

Given mass of ethane = 15.42 g

Molar mass of ethane = 30.07 g/mol

Putting values in above equation, we get:

$\text{Moles of ethane}=\frac{15.42g}{30.07g/mol}=0.513mol$

The chemical equation for the combustion of ethane follows:

$2C_2H_6+5O_2\rightarrow 4CO2+6H_2O$

By Stoichiometry of the reaction:

2 moles of ethane produces 6 moles of water

So, 0.513 moles of ethane will produce = $\frac{6}{2}\times 0.513=1.54mol$ of water

Hence, the moles of water produced are 1.54 moles.

4 0

3 years ago

Which of the following sets of oxidation numbers is NOT correct? KMnO4 : K +2, Mn +6, O -2 NaCHO2 : Na +1, C +2, H +1, O -2 Zn(O

muminat

KMnO4 has the incorrect set of oxidation numbers. Whenever there is an alkali metal, it has an oxidation number of +1. As you can see, K is said to have an oxidation number of +2, so it is incorrect.

6 0

3 years ago

A student uses a calorimeter to determine the enthalpy of dissolving for ammonium nitrate. The student fills a calorimeter with

ale4655 [162]

Enthalpy change during the dissolution process = m c ΔT,

here, m = total mass = 475 + 125 = 600 g
c = specific heat of water = 4.18 J/g °C
ΔT = 7.8 - 24 = -16.2 oc (negative sign indicates that temp. has decreases)

Therefore, Enthalpy change during the dissolution = 600 x 4.18 X (-16.2)
= -40630 kJ
(Negative sign indicates that process is endothermic in nature i.e. heat is taken by the system)

Thus, enthalpy of dissolving of the ammonium nitrate is -40630 J/g

7 0

3 years ago

A sheet of polyethylene 1.5-mm thick is being used as an oxygen barrier at a temperature of 600K. If the flux is 2.48 x 10-5 kg/

Eddi Din [679]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of high-pressure side is $C_1 = 0.722 \ kg/m^3$