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3 years ago

If a molecule has an empirical formula of C2H2O and a molecular mass of 84.0 g/mol, what is the molecular formula?

Chemistry

2 answers:

Ksivusya [100]3 years ago

7 0

Answer:

$C_2 H_2 O \times 2=C_4 H_4 O_2$ is the molecular formula

Explanation:

Empirical formula is the simplest form of a Molecular formula.

For example

Molecular formula of glucose is $C_6 H_{12} O_6$

Its empirical formula is $C_1 H_2 O_1$

The subscript numbers of the molecular formula is divided by a common number n=6 here.

Another example:

Molecular formula of Octane is $C_8 H_{18}$ and Its empirical formula will be $C_4 H_9$

Subscripts divided by n=2.

To find n we make use of the formula

$\frac {(molecular mass )}{(empirical formula mass)}=n$

$= \frac {84.0g}{(C_2H_2O mass)}\\\\=\frac {84.0g}{(24+2+16)g}=\frac {84.0g}{42g}=2$

So, we see n=2 here

Empirical formula × n = molecular formula

$C_2 H_2 O \times 2=C_4 H_4 O_2$ is the molecular formula

(Answer)

Please note:

Molar mass is the mass of 1 mole of the substance and its unit is g/mol .

n is the number of moles by which the empirical formula is multiplied to get the molecular formula of the compound.

lord [1]3 years ago

6 0

Answer:

=C₄H₄O₂

Explanation:

Given the empirical formula of a molecule, the he the quotient of the molecular mas and and the empirical mass=constant.

84.0 g/mol/mass of(C₂H₂O)=constant

=84/(12×2+1×2×16)

=84/42

Therefore, the molecular formula is (C₂H₂O)₂=C₄H₄O₂

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Commercially available solution of SPT has a density of 3.10 g/mL. Water (1.00 g/mL) can be added to the original SPT solution,

Pepsi [2]

The reciprocal of the density, ρ, of a mixture is equal to the sum of the reciprocal of the densities of the constituents multiplied by their respective proportions ( $x_i$ ) in the solution, $\displaystyle \dfrac{1}{\rho_{(mixture)}} = \sum \dfrac{x_i}{\rho _i}$

The percentage of water in the 2.75 g/mL SPT solution is $\underline{16. \bar6 \ \%}$

Reason:

Known parameter:

Density of SPT = 3.10 g/mL

Density of water = 1.00 g/mL

Required:

The percentage of water in 2.75 g/mL SPT solution

Solution:

Let x the volume of water added per 100 mL of 2.75 g/mL SPT solution, we have;

Mass of water in the solution = x mL × 1.00 g/mL = x g

Mass of commercial SPT = (100 - x) mL × 3.10 g/mL = (310 - 3.1·x) g

$Density = \dfrac{Mass}{Volume}$

The density of the 2.75 g/mL SPT solution is given as follows;

$Density \ of \ 100\ mL, \ 2.75 \ g/mL = \dfrac{(310 - 3.1\cdot x + x) \ g}{100 \ mL}$

100 mL × 2.75 g/mL =( 310 - 2.1·x) g

310 - 275 = 2.1·x

$x = \dfrac{310 - 275}{2.1} = \dfrac{50}{3} = 16.\overline 6$

Given that the reference quantity of the proportion of water in the solution is 100 mL of the 2.75 g/mL SPT solution, the percentage (per 100) of water in the overall solution is the value of x

Therefore;

The percent of the SPT solution that must consist of water to give an overall density of 2.75 g/mL is x = $\displaystyle {\underline {16.\overline 6 \ \%} }$