Question

Рассчитайте молекулярные массы(Mr) следующих веществ:а)глюкоза (С6H12О6)б)глицерин(C3H8O3)в)этиловый эфир муравьиной кислоты(C3H6O2)

Answer 1

Mr а)(12*6+1*12+16*6=180
б)12*3+1*8+16*3=92
в)12*3+1*6+16*2=74

Answer 2

Calculation of Molecular Masses (Mr) for the Given Substances

To calculate the molecular masses (Mr) of the given substances, we need to determine the sum of the atomic masses of all the atoms in each molecule.

a) Glucose (C6H12O6): - Carbon (C) atomic mass = 12.01 g/mol - Hydrogen (H) atomic mass = 1.008 g/mol - Oxygen (O) atomic mass = 16.00 g/mol

The molecular mass (Mr) of glucose can be calculated as follows: (6 * C) + (12 * H) + (6 * O) = (6 * 12.01) + (12 * 1.008) + (6 * 16.00) = 180.18 g/mol

Therefore, the molecular mass (Mr) of glucose (C6H12O6) is 180.18 g/mol.

b) Glycerin (C3H8O3): - Carbon (C) atomic mass = 12.01 g/mol - Hydrogen (H) atomic mass = 1.008 g/mol - Oxygen (O) atomic mass = 16.00 g/mol

The molecular mass (Mr) of glycerin can be calculated as follows: (3 * C) + (8 * H) + (3 * O) = (3 * 12.01) + (8 * 1.008) + (3 * 16.00) = 92.09 g/mol

Therefore, the molecular mass (Mr) of glycerin (C3H8O3) is 92.09 g/mol.

c) Ethyl Formate (C3H6O2): - Carbon (C) atomic mass = 12.01 g/mol - Hydrogen (H) atomic mass = 1.008 g/mol - Oxygen (O) atomic mass = 16.00 g/mol

The molecular mass (Mr) of ethyl formate can be calculated as follows: (3 * C) + (6 * H) + (2 * O) = (3 * 12.01) + (6 * 1.008) + (2 * 16.00) = 74.08 g/mol

Therefore, the molecular mass (Mr) of ethyl formate (C3H6O2) is 74.08 g/mol.

Please let me know if there is anything else I can help you with!