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Reptile [31]
1 year ago
12

Vinegar contains carbon, hydrogen, and oxygen with percent masses of 40.01% C, 6.70% H, and 53.29% O, respectively. Determine th

e empirical and molecular formulas of the compound. The molar mass of vinegar is about 60 g/mol. HOW DO WE GET THERE
Chemistry
1 answer:
kobusy [5.1K]1 year ago
3 0

The empirical formula is = C2H4O2

Molecular formula=C2H4O2

<h3>Calculation of Molecular formula and empirical formula:-</h3>

Vinegar has three elements:

oxygen = 53.29%,

hydrogen = 6.70%,

and carbon = 40.01%.

We can create an equation for the total mass of vinegar with 'a' carbon atom,' b' hydrogen atom, and 'c' oxygen atoms using the molar masses of C, H, and O.

12*a + 1*b + 16*c = 60

We also know that C makes up 24 g/mol, or 40.01 percent, of the overall mass of 60 g/mol.

The formula contains two C atoms because each C atom has a molecular mass of 12 g/mol. You may determine that the formula contains 4 H and 2 O atoms by using the same reasoning for H and O.

The empirical formula is = C2H4O2

Molecular formula= (C2H4O2)n

(4 x 12 + 1 x 4 + 16 x 2)n = 60

(84)n = 60

n=60/84

n=0.71=1

Molecular formula=C2H4O2

Learn more about Molecular and empirical formulas here:-

brainly.com/question/9207476

#SPJ4

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3 years ago
Which scale is being described? Water freezes at 32. Water freezes at 0. Water freezes at 273.
Leya [2.2K]

Scale is defined as the measurement of unit ranges, that has been corresponds to the measurements.

The units for the measurement of temperature are given as:

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The unit scale for the measurement of freezing of water are

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7 0
2 years ago
Identify the correct coefificients to balance it<br> -C3H8+O2 to -CO2 +-H2O
Iteru [2.4K]

Answer:

{\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

Explanation:

{\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}.

Among the four species in this reaction, \rm C_3H_8 is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is 1.

{\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}.

Number of atoms on the left-hand side of the reaction:

  • \rm C: 1 \times 3 = 3.
  • \rm H: 1 \times 8 = 8.
  • \rm O: not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, \rm CO_2 is the only product with carbon atoms, whereas \rm H_2O is the only product with hydrogen atoms. These 3 carbon atoms and 8 hydrogen atoms would correspond to:

  • 3 / 1 = 3 \rm CO_2 molecules, and
  • 8 / 2 = 4 \rm H_2O molecules.

{\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

Number of atoms on the right-hand side of the reaction:

  • \rm C: 3 \times 1 = 3.
  • \rm H: 4 \times 2 = 8.
  • \rm O: 3 \times 2 + 4 \times 1 = 10.

The number of \rm O atoms on the left-hand side should match those on the right-hand side. In this reaction, \rm O_2 is the only reactant with \rm O\! atoms. These 10 \rm \! O atoms would correspond to:

  • 5 \rm O_2 molecules.

{\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

5 0
2 years ago
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