Question

Calculate the % composition of oxygen in MgSO4.7H2O​

Answer 1

Answer:

Relative molecular mass of MgSO 4.7H 2 O

=24+32+(16×4)+7(2+16)

=24+32+64+126=246

26g of Epsom salt contains 126g of water of crystallisation.

Hence, 100g of Epsom salt contains 100×126/246

The % of H 2 O in MgSO 4 .7H2O =51.2

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Answer 2

The percentage composition of oxygen $MgSO_{4} .7H_{2}O$ is $51.2$.

Step-by-step explanation:

Given:

The given Magnesium sulfate heptahydrate is $MgSO_{4} .7H_{2}O$.

To find:

To find the % composition of oxygen in $MgSO_{4} .7H_{2}O$.

Solution:

The molecular mass of $MgSO_{4} .7H_{2}O=$the atomic mass of $Mg+$ atomic mass of $S+4\times$ atomic mass $O+7(2\times$ atomic mass of $H+$atomic mass of$O)$.

$=24+32+4\times 16+7(2\times1+16)g/mol$

$=(24+32+62+126)g/mol$

$=246g/mol$

Molecular mass of total water $=7(2\times$ atomic mass of $H+$ atomic mass of $O)$.

$=7\times18=126g/mol$

Now calculate the percentage of oxygen in EPSOM salt.

Percentage of oxygen in EPSOM salt = total molar mass of $H_{2}O$/ molar mass of EPSOM salt $\times 100$.

$=126/246\times100$

$=12600/246$

$=51.2$

Hence,  the % composition of oxygen in $MgSO_{4} .7H_{2}O$ is $51.2$ percentage.

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