A compound is found to be 38.76% calcium, 19.97% phosphorous and 41.28% oxygen. What is the empirical formula for this compound?
1 answer:
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Answer:
Explanation:
Density is m/V. Also, 1 liter = 1000 . So, we get 0.890/(5*1000) =
g/cm^3. You can convert this to kg/m^3 as well by multiplying it by 10. Depends which one you want.
Answer:
Explanation:
We want to convert from moles to grams, so we must use the molar mass.
<h3>1. Molar Mass</h3>The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).
We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.
- Phosphorus (P): 30.973762 g/mol
- Iodine (I): 126.9045 g/mol
Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.
- I₃: 126.9045 * 3=380.7135 g/mol
- PI₃: 30.973762 + 380.7135 = 411.687262 g/mol
Use the molar mass as a ratio.
We want to convert 3.14 moles to grams, so we multiply by that value.
The units of moles of PI₃ cancel.
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.
- 1292.698
The 2 in the ones place tells us to leave the 9.
3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>