Respuesta :
Answer:
The answer to your question is Scaling factor = 5
Explanation:
Data
Empirical formula CH₂O
molar mass = 150 g/mol
Empirical formula is a chemical formula that shows the simplest proportions of the elements of that formula.
Process
1.- Calculate the molar mass of the Empirical formula
CH₂O = (12 x 1) + (2 x 1) + (16 x 1)
= 12 + 2 + 16
= 30 g
2.- Divide the molar mass of the molecular formula by the molar mass of the empirical formula.
150g / 30g = 5 Scaling factor = 5
3.- Molecular formula
5(CH₂O) = C₅H₁₀O₅
The scaling factor is 5
Considering the question given above, the following data were obtained:
Empirical formula = CH₂O
Molar mass of compound = 150 g/mol
Scaling factor (n) =?
The scaling factor can be obtained as follow:
Empirical formula × n = molar mass
[CH₂O]n = 150
[12 + (2×1) + 16]n = 150
[12 + 2 + 16]n = 150
30n = 150
Divide both side by 30
[tex]n = \frac{150}{30}\\\\[/tex]
n = 5
Therefore, the scaling factor is 5
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