(ii) Here, oxgen acts as a limiting reagent as only 16 g of dioxygen is available. Therefore 1 mole of carbon dioxide produced by burning 1 mole of carbon. The balanced reaction of combustion of carbon in dioxygens is: (iii) 2 moles of carbon are burnt in 16 g of dioxygen. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. Calculate the amount of carbon dioxide that could be produced when ∴ The empirical formula of the iron oxide is Fe 2O 3.Ĥ. Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88 Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25 Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass. Mass percent of an element = (Mass of that element in compound/Molar mass of that compound) × 100 Calculate the mass percent of different elements present in Sodium Sulphate (Na 2SO 4). = (1 x Atomic weight of carbon) + (2 x Atomic weight of oxygen)Ģ.

= Molecular weight of carbon dioxide, CO 2 = (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen) = (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen) Calculate the molecular mass of the following : Our Physics Wallah team Prepared Other Subjects NCERT Solutions for class 11.Ĭhapter 1 Some Basic Concepts of Chemistryġ. NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry is prepared by our senior and renown teachers of Physics Wallah primary focus while solving these questions of class-11 in NCERT textbook, also do read theory of this Chapter 1 Some Basic Concepts of Chemistry while going before solving the NCERT questions. NCERT Solutions For Class 11 Chemistry chapter 1-Some Basic Concepts of Chemistry

Carbon dioxide formula mass