Question #533df

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Answer 1 · 2015-10-08T00:22:40

$M g S O_{4} .7 H_{2} O$ $MgSO_(4).7H_2O$

Explanation:

The hydrated magnesium sulfate contains water of crystallisation which is driven off when heated:

$M g S O_{4} . x H_{2} O_{(s)} \to M g S O_{4 (s)} + x H_{2} O_{(g)}$ $MgSO_(4).xH_2O_((s))rarrMgSO_(4(s))+xH_2O_((g))$

By recording the loss in mass we can find the mass of water and then convert masses into moles to find $x$ $x$ .

Mass $M g S O_{4} . x H_{2} O = 4.312 g$ $MgSO_(4).xH_2O=4.312"g"$

Mass $M g S O_{4} = 2.107 g$ $MgSO_(4)=2.107"g"$

So mass water lost = $4.312 - 2.107 = 2.205 g$ $4.312-2.107=2.205"g"$

Now we need to convert grams into moles using:

Number of moles = mass in grams/mass of 1 mole

Or:

$n = \frac{m}{M_{r}} (1)$ $n=m/M_r " "color(red)((1))$

$M_{r}$ $M_r$ is the relative molecular mass for a molecular compound or the relative formula mass for an ionic compound.

We get this by adding all the relative atomic masses ( $A_{r}$ $A_r$ ) together as they occur in the formula. You should always be provided with these in the question or on a data sheet.

Using accurate $A_{r}$ $A_r$ values we get:

$M_{r} [M g S O_{4}] = 120.37$ $M_r[MgSO_4]=120.37$

$M_{r} [H_{2} O] = 18.01$ $M_r[H_2O]=18.01$

You should use the $A_{r}' s$ $A_r's$ you are given - they will be close.

Now using $(1) \Rightarrow$ $color(red)((1))rArr$

$n M g S O_{4} = \frac{2.107}{120.37} = 0.0175$ $nMgSO_4=2.107/120.37=0.0175$

and:

$n H_{2} O = \frac{2.205}{18.01} = 0.12243$ $nH_2O=2.205/18.01=0.12243$

So we now know that the ratio in moles of:

$M g S O_{4} : H_{2} O$ $MgSO_4:H_2O$ is:

$0.0175 : 0.12243$ $0.0175:0.12243$

We like to express this ratio in whole numbers according to how the atoms/molecules combine so we can divide both numbers by $0.0175 \Rightarrow$ $0.0175rArr$

$\frac{0.0175}{0.0175} : \frac{0.12243}{0.0175} \Rightarrow$ $0.0175/0.0175:0.12243/0.0175rArr$

$1 : 6.996$ $1:6.996$

Which is good enough (in my book) as:

$1 : 7$ $1:7$

So $x = 7$ $x=7$ which gives the formula for hydrated magnesium sulfate to be:

$M g S O_{4} .7 H_{2} O$ $MgSO_(4).7H_2O$

Here is a full lab on hydrate and how to calculate the formula of a hydrate, I hope this helps more:

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