Chemistry, 06.07.2019 22:30 AshlynPlayz45

Acompound is found to be 38.76% calcium, 19.97% phosphorous and 41.28% oxygen. what is the empirical formula for this compound?

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Answer from: deidrabenson72331

The empirical formula for this compound in the given case would be - Ca₃P₂O₈ or Ca_3(PO_4)_2 .

Let assume the given compound is 100g than the molar quantities of each element.

=> Moles of calcium

= 0.967 mol

=> Moles of phosphorus

= $\frac{19.97g}{31.0g}$

= 0.644 mol

=> Moles of oxygen

= 2.58 mol

Multiply integers of molecules by a number that makes them close to integers.

Element Mass/g Moles Ratio ×2 Integers

Ca 38.76 0.967 07 1.4998 2.9995 3

P 19.97 0.644 82 1 2 2

O 41.28 2.580 0 4.0011 8.0023 8

Thus, the empirical formula is Ca₃P₂O₈.

Learn more about the empirical formula:

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Answer from: krojas015

The empirical formula is Ca₃P₂O₈.

Assume that you have 100 g of the compound.

Then you have 38.76 g Ca, 19.97 g P, and 41.28 g O.

Now, we must convert these masses to moles and find their ratios.

If the number in the ratio are not close to integers, you multiply them by a number that makes them close to integers.

From here on, I like to summarize the calculations in a table.

Element Mass/g Moles Ratio ×2 Integers

Ca 38.76 0.967 07 1.4998 2.9995 3

P 19.97 0.644 82 1 2 2

O 41.28 2.580 0 4.0011 8.0023 8

The empirical formula is Ca₃P₂O₈.

Answer from: Quest

we need the questions , i may then you.

Answer from: Quest

Iwould you figure it out but you don't have a picture or in your question as to what the physical property you're looking for is

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Acompound is found to be 38.76% calcium, 19.97% phosphorous and 41.28% oxygen. what is the empirical...

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