Question

A 35.40 gram hydrate of sodium carbonate, Na2CO3•nH2O, is heated to a constant mass. Its final weight is 30.2 g. What is formula for the hydrate?

Answer 1

Answer:

Na₂CO₃•H₂O

Explanation:

After it is heated, the remaining mass is the mass of sodium carbonate.

30.2 g Na₂CO₃

Mass is conserved, so the difference is the mass of the water:

35.4 g − 30.2 g = 5.2 g H₂O

Convert masses to moles:

30.2 g Na₂CO₃ × (1 mol Na₂CO₃ / 106 g Na₂CO₃) = 0.285 mol Na₂CO₃

5.2 g H₂O × (1 mol H₂O / 18.0 g H₂O) = 0.289 mol H₂O

Normalize by dividing by the smallest:

0.285 / 0.285 = 1.00 mol Na₂CO₃

0.289 / 0.285 = 1.01 mol H₂O

The ratio is approximately 1:1.  So the formula of the hydrate is Na₂CO₃•H₂O.

Answer 2

Na₂CO₃•H₂O

Further explanation

Given:

• A 35.40 g hydrate of sodium carbonate, Na₂CO₃•nH₂O, is heated to a constant mass.
• Its final weight is 30.2 g.

Question:

What is the formula for the hydrate?

Problem-solving:

We will solve problems related to The Law of Definite Proportion (Proust's Law).

"The ratio of the masses of elements in each compound is always constant. Put differently, a given compound invariably contains a similar proportion of elements by mass."

The heating reaction of the hydrate of sodium carbonate is as follows:

$\boxed{ \ Na_2CO_3.nH_2O \xrightarrow{\text{heat}} Na_2CO_3 + nH_2O \ }$

Condition:

• Before heating: 35.40 g of Na₂CO₃•nH₂O (hydrate)
• After heating: 30.20 g of Na₂CO₃ (anhydrate)
• The mass of H₂O that has left the compound is $\boxed{ \ 35.40 - 30.20 = 5.2 \ g \ }$  

$\boxed{ \ Na_2CO_3.nH_2O \xrightarrow{\text{heat}} Na_2CO_3 + nH_2O \ }$

         35.4 g                 30.2 g       5.2 g

We want to determine the amount of water (with the symbol n) so that the formula for the hydrate can be known.

Recall that $\boxed{ \ Moles = \frac{Mass \ (g)}{Mr} \ }$

Let us prepare the number of moles of Na₂CO₃ and H₂O respectively.

• Mr Na₂CO₃ = 106 g/mol ⇒ $\boxed{ \ moles = \frac{30.2}{106} \rightarrow \boxed{ \ 0.285 \ } \ }$
• Mr H₂O = 18 g/mol ⇒ $\boxed{ \ moles = \frac{5.2}{18} \rightarrow \boxed{ \ 0.289 \ }\ }$

Finally, we use the mole ratio to write the formula. Find the water-to-anhydrate mole ratio.

$\boxed{ \ Na_2CO_3 \ : \ H_2O = 1 \ : \ n  \ }$

$\boxed{ \ H_2O \ : \ Na_2CO_3 = n \ : \ 1  \ }$

$\boxed{ \ n \ : \ 1 = 0.289 \ : \ 0.285 \ }$

$\boxed{ \ n = \frac{0.289}{0.285} \times 1 \ }$

Thus, the value of n after rounding is $\boxed{ \ n = 1 \ }$

Substitute n = 1 into Na₂CO₃•nH₂O.

Therefore, the formula for the hydrate is $\boxed{\boxed{ \ Na_2CO_3.H_2O \ }}$

Learn more

1. The law of multiple proportions (Dalton’s law)  https://brainly.com/question/10590259  
2. Conservation of mass  https://brainly.com/question/9473007
3. The chemical formula of a compound  https://brainly.com/question/834909

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