Find the mass of C in the 2.657 g CO2:
(2.657 g CO2) / (44.01 g/mol) = 0.06037 mol CO2
Since each mole of CO2 also has 1 mole of C, this is equivalent to 0.06037 mol C.
Find the mass of H in the 1.089 g H2O:
(1.089 g H2O) / (18.02 g/mol) = 0.06043 mol H2O
Since 1 mol H2O has 2 mol H, this is equivalent to (0.06043)*2 = 0.1209 mol H.
Taking the ratio of H to C: 0.1209 / 0.06037 = 2.002 ~ 2
Therefore, the empirical formula of isobutylene is CH2.
I would say you should use or test it once a week to ensure it is working properly in an active laboratory since it is a workplace with significant chemical hazards so it would give peace of mind to know on a quite regular basis that it can be relied on in case of an emergency like an eye flush for example.
name= calcium
atomic mass= 40.078 atomic mass unit
no of protons= 20
no of electrons=20
The answer is letter (B).
