1.145 Sou 5, Asample of phosphorus weighing 0.500 g was ignited to the oxide in

Question

1.145 Sou 5, Asample of phosphorus weighing 0.500 g was ignited to the oxide in a stream of oxygen A1 41 gas. What is the empirical formula of phosphorus oxide if the product has a mass of 1.145 g? What is the molecular formula of phosphorus oxide if the molar mass is approximately 285 g/mol? 6,00 Empirical formula Molecular formula 6 (optional) Ethylene glycol, the main ingredient in antifreeze, contains 38.7% carbon, 9.7% hydrogen, and 51.6% oxygen. Calculate the empirical and molecular formulas for ethylene glycol, given the molar mass is approximately 60 g/mol. Empirical formula Molecular formula 166

Explanation / Answer

5)

we have mass of each elements as:

P: 0.500 g

O: 1.145 - 0.500 = 0.645 g

Divide by molar mass to get number of moles of each:

P: 0.5/30.97 = 0.0161

O: 0.645/16.0 = 0.0403

Divide by smallest to get simplest whole number ratio:

P: 0.0161/0.0161 = 1

O: 0.0403/0.0161 = 2.5

So empirical formula is:

P2O5

Molar mass of P2O5,

MM = 2*MM(P) + 5*MM(O)

= 2*30.97 + 5*16.0

= 141.94 g/mol

Now we have:

Molar mass = 285.0 g/mol

Empirical formula mass = 141.94 g/mol

Multiplying factor = molar mass / empirical formula mass

= 285.0/141.94

= 2

Hence the molecular formula is : P4O10

Answer:

Empirical formula: P2O5

Molecular formula: P4O10

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