The compound tyrosine contains 59.66% c, 6.12% h, 7.73% n, and 26.49% o by mass. what is the empirical formula of tyrosine?

C9H11O3N

Explanation:

The percentage by mass of each element is divided by its relative atomic mass. The lowest ratio is now observed. Each ratio is now divided by this lowest ratio and approximated to whole numbers. All these are shown in detail in the image attached.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g