Answered

Connect with a global community of knowledgeable individuals on IDNLearn.com. Get the information you need from our community of experts who provide accurate and comprehensive answers to all your questions.

the molecular formula mass of this compound is 150 amu . what are the subscripts in the actual molecular formula?

Sagot :

Louanaalbert676idn

Answer:

You need the mass composition of the compound.

This composition works fine for the formula mass 150 amu.

C: 40.0%

H: 6.7%

O: 53.3%

From that, you can solve the problem following theses steps:

1) Convert mass composition fo molar ratios by dividing ech element by its atomic mass =>

C: 40.0 / 12 = 3.33

H: 6.70 / 1 = 6.70

O: 53.3 / 16 = 3.33

2) Divide all the numbers by the smallest one =>

C: 3.33 / 3.33 = 1

H: 6.70 / 3.33 = 2

O: 3.33 / 3.33 = 1

3) Write the empirial formula and calculate its mass:

C H2 O => 1 * 12.0 g/mol + 2 * 1.0 g/mol + 1 * 16.0 g/mol = 30 g/mol

4) Calculate how many times the empirical mass is contained in the molecular mass:

150 / 30 = 5

5) Conclusion:

The molecular formula is C5 H10 O5, i.e. the number of each atom in the molecular formula are:

C =5, H = 10, O = 5

We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for solutions ends here at IDNLearn.com. Thank you for visiting, and come back soon for more helpful information.