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Answer 1 · 2017-06-23T17:20:45

$X = 32.1 \frac{g}{m}$ $X = 32.1 g/m$

Set up a ratio and solve for X

$\frac{5.72}{10} = \frac{4 x}{4 x + 6 (O)}$ $5.72 /10 = (4x)/ ( 4x + 6(O)$

5.72 = the amount of X in the compound

4x = the amount of X in the compound

so $5.72 = 4 x$ $5.72 = 4x$

10 = the total amount of the Compound

4x + 6(O) the total amount of the Compound

so $10 = 4 X + 6 (O)$ $10 = 4X + 6(O)$ The ratio is

$\frac{5.72}{10} = \frac{4 X}{4 X + 6 O}$ $5.72/10 = (4X) / ( 4X + 6O)$

$6 \times 16 = 96$ $6 xx 16 = 96$ : $O = 16 \frac{g}{m}$ $O = 16g/m$

$\frac{5.72}{10} = \frac{4 x}{4 X + 96}$ $5.72/10 = (4x)/(4X + 96)$ removing the denominators gives

$5.72 (4 x) + 5.72 (96) = 40 X$ $5.72(4x) + 5.72(96) = 40X$ multiplying this out gives.

$22.88 X + 549 . = 40 X$ $22.88 X + 549. = 40X$ subtract 22.88 X from both sides

$22.88 X - 22.88 X + 549 = 40 X - 22.88 X$ $22.88 X - 22.88 X + 549 = 40X - 22.88X$ This gives

$549 = 17.12 X$ $549 = 17.12 X$ divide both sides by 17.12 gives

$32.1 g . = X$ $32.1 g. = X$

Within the margin of error the grams / mole is approximately 32 grams/mole X is probably Sulfur