What is the molar mass of $H_{3} P O_{4}$ $H_3PO_4$ ?

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Answer 1 · 2016-03-31T20:48:29

$97.995 g / m o l$ $97.995g//mol$

You need to add the atomic masses of the molecule:

$M (H_{3} P O_{4}) = 3 \times M (H) + M (P) + 4 \times M (O)$ $M(H_3PO_4)=3xxM(H)+M(P)+4xxM(O)$

$= 3 \times 1.008 + 30.9738 + 4 \times 15.9994 = 97.995 g / m o l$ $=3xx1.008+30.9738+4xx15.9994=97.995g//mol$

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