The percent composition of all elements in HClO3 is: a. 2% H, 20% Cl, 78% O b. 1% H, 42% Cl, 57% O c. 2% H, 68% Cl, 30% O d. None of the above
well take molar mass of HClO3 and atomic masses of atoms in compound and divide each atomic mass with molar mass of compound
idk how to do that, im completelylost because it has different elements ombined.
H - 1,008 Cl - 35,45 O - 15,999 M(HClO3) = (3* 15,999) + 1,008 + 35,45 = 84,455 x(H) = 1,008/84,455 =0,0119 = 1,2 % x(Cl) = 35,45/ 84,455= 0,419 = 42% x(O)= 3*15,999 / 84,455=0,5683 = 57 %
so answer is: b. 1% H, 42% Cl, 57% O
thank you so muchchemistry is so confusing. im still in the beginning of this section with mole calculations and i still dont understand it.
read it few times so it settles down in your head, after you learn basics chemistry is pretty easy... :)
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