a compound has an empirical formula of C3H5O2 and a molar mass of 145,14g. what is the molecular formula of this compound?

empirical formula mass = 3C+5H+2O =

3*12+5*1+2*16 = 36+5+32= 73.

So 145.14/73 = 1.988. Round this to an even number (in this case 2.0) and the molecular formula is
(C3H5O2)2 or C6H10O4. That is, the molecular formula is two units of the empirical formula.

Pls i need the answer

calculate the ph solution with hydrogen ion concentration of 6.28×10^6 moldm and 1×10^-5

To determine the molecular formula of a compound using the empirical formula and molar mass, you need to find the ratio between the empirical formula mass and the molar mass. Once you have this ratio, you can multiply the subscripts of the empirical formula by this value to find the molecular formula.

Step-by-step process:

1. Calculate the empirical formula mass:
- The empirical formula has 3 carbon (C) atoms with a molar mass of 12.01 g/mol each, so the total mass contributed by carbon is 3 * 12.01 g/mol = 36.03 g/mol.
- The empirical formula has 5 hydrogen (H) atoms with a molar mass of 1.01 g/mol each, so the total mass contributed by hydrogen is 5 * 1.01 g/mol = 5.05 g/mol.
- The empirical formula has 2 oxygen (O) atoms with a molar mass of 16.00 g/mol each, so the total mass contributed by oxygen is 2 * 16.00 g/mol = 32.00 g/mol.
- Add up the masses contributed by each element: 36.03 g/mol + 5.05 g/mol + 32.00 g/mol = 73.08 g/mol.

2. Calculate the ratio between the empirical formula mass and the molar mass:
- Divide the molar mass of the compound (145.14 g/mol) by the empirical formula mass (73.08 g/mol):
145.14 g/mol ÷ 73.08 g/mol ≈ 1.986 (rounded to three decimal places).

3. Multiply the subscripts in the empirical formula by the ratio calculated in Step 2:
- Multiply the subscripts for each element in the empirical formula (C3H5O2) by 1.986:
C3H5O2 * 1.986 ≈ C5.958H9.93O3.972 (rounded to three decimal places).

4. Round the resulting subscripts to the nearest whole number to derive the molecular formula:
- The molecular formula based on the rounded subscripts becomes C6H10O4.

Therefore, the molecular formula of the compound is C6H10O4.

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