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A compound has a percent composition of 15.24% sodium (molar mass = 22.99
g/mol), 52.95% bromine (molar mass = 79.90 g/mol), and 31.81% oxygen (molar
mass = 16.00 g/mol). Assuming that the mass of the compound is 100 g, what is
the compound's empirical formula?
Show all your work. Please use correct formatting for subscripts and exponents. The math for-

Respuesta :

Eduard22sly

The empirical formula of the compound obtained from the question given is NaBrO₃

Data obtained from the question

  • Sodium (Na) = 15.24%
  • Bromine (Br) = 52.95%
  • Oxygen (O) = 31.81%
  • Empirical formula =?

How to determine the empirical formula

The empirical formula of the compound can be obtained as illustrated below:

Divide by their molar mass

Na = 15.24 / 22.99 = 0.663

Br = 52.95 / 79.90 = 0.663

O = 31.81 / 16 = 1.988

Divide by the smallest

Na = 0.663 / 0.663 = 1

Br = 0.663 / 0.663 = 1

O = 1.988 / 0.663 = 3

Thus, the empirical formula of the compound is NaBrO₃

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