A compound composed of 3. 3% h, 19. 3% c, and 77. 4% o has a molar mass of approximately 60 g/mol. What is the molecular formula of the compound?.

Respuesta :

The molecular of the compound will be-(H₂CO₃)1= H₂CO₃

The given molecule has the molecular formula H2CO3, also known as Carbonic acid.

Given that-

Consequently,

H= 3.3%.

C=19.3%

O =77.4%

Number of H moles = 3.3/1 = 3.3

Number of C moles = 19.3/12 = 1.60

Number of O moles = 77.4/16 = 4.83

As a result, the ratio of C, H, and O atoms is 3.3: 1.60: 4.83.

Divide by the amount you obtain that is the smallest: 3.3/1.60; 1.60/1.60; 4.83/1.60 = 2: 1: 3

H2CO3 is the empirical formula.

Let (H2CO3)n be the molecular formula.

A molecular formula is a chemical formula that specifies how many atoms of each element there are in each molecule of a given substance.

Molar mass thus equals (21+112+3X16)n = 62n.

As per the inquiry, 62n = 60

alternatively, n= 0.96 (round figure is 1)

Therefore, carbonic acid is the chemical whose molecular formula is (H2CO3)1=H2CO3.

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