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Lady bird [3.3K]
3 years ago
15

What is the empirical formula for a compound that contains 79.86 % iodine and 20.14 % oxygen by mass?

Chemistry
1 answer:
serg [7]3 years ago
7 0

Answer:

IO₂

Explanation:

We have been given the mass percentages of the elements that makes up the compound:

Mass percentage given are:

Iodine = 79.86%

Oxygen = 20.14%

To calculate the empirical formula which is the simplest formula of the compound, we follow these steps:

> Express the mass percentages as the mass of the elements of the compound.

> Find the number of moles by dividing through by the atomic masses

> Divide by the smallest and either approximate to nearest whole number or multiply through by a factor.

> The ratio is the empirical formula of the compound.

Solution:

I O

% of elements 79.86 20.14

Mass (in g) 79.86 20.14

Moles(divide by

Atomic mass) 79.86/127 20.14/16

Moles 0.634 1.259

Dividing by

Smallest 0.634/0.634 1.259/0.634

1 2

The empirical formula is IO₂

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If you combine Sodium and Water, chemical changes causes decomposition into Sodium Hydroxide and Hydrogen.

Sodium + Water  ==>  Sodium Hydroxide and Hydrogen

Na + H2O ====>  NaOH and H

Another example of chemical change is:

Carbon Dioxide and Water will decompose into Sugar and Oxygen

Carbon Dioxide + Water ==> Sugar and Oxygen

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7 0
3 years ago
What is the formula for the compound dinitrogen pentoxide? n4o5 n5o4 n4o6 n5o2 n2o5?
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5 0
3 years ago
Read 2 more answers
how many grams of calcium sulfate are produced from 10 grams of calcium nitrate and how many grams of calcium sulfate are produc
AlekseyPX

Answer:  8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Ca(NO_3)_2=\frac{10g}{164g/mol}=0.061moles

\text{Moles of} Li_2SO_4=\frac{10g}{110g/mol}=0.091moles

Ca(NO_3)_2+Li_2SO_4\rightarrow 2LiNO_3+CaSO_4  

According to stoichiometry :

1 mole of Ca(NO_3)_2 require = 1 mole of Li_2SO_4

Thus 0.061 moles of Ca(NO_3)_2 will require=\frac{1}{1}\times 0.061=0.061moles  of Li_2SO_4

Thus Ca(NO_3)_2 is the limiting reagent as it limits the formation of product and Li_2SO_4 is the excess reagent.

As 1 mole of Ca(NO_3)_2 give = 1 mole of CaSO_4

Thus 0.061 moles of Ca(NO_3)_2 give =\frac{1}{1}\times 0.061=0.061moles  of CaSO_4

Mass of CaSO_4=moles\times {\text {Molar mass}}=0.061moles\times 136g/mol=8.30g

Thus 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

6 0
3 years ago
An aqueous solution of glucose has molality of 6.81 and a density of 1.2 g/ml what is the molarity of glucose in the solution
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Answer:

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the molality of the solution of sucrose is

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M= 3.68 moles/L

Note:

- Would be wrong in this case to assume density of water = 1 Kg/L since the solution is heavily concentrated in glucose and therefore the density of water deviates from its pure value.

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