Question
Guy Barnes
Arbitrator

# Weight of Sand, Cement and Water for Mortar Mix Ratio 1:3

How to calculate exactly the weight of sand cement and water in mix ratio 1:3?

### Where am I?

Quantity is calculated for 1 Cu.m of mortar:

For 1 Cu.m mortar

Cement Sand W/C ratio Remarks

1 3 0.5

Total 4 =1+3

For 1 Cu.m

Considering 10% Wastage

Total Quantity of wet mortar 1.1 10% of 1 Cu.m

Shrinkage factor 1.25

Total Quantity of dry mortar (Wet mortar x Shrinkage factor) 1.375 Multiplying with wet mortar volume=1.25*1.1

Cement

Cement 0.34 =(1/4)(1.375)

Cement Bag 9.96 =0.375/0.0345 (0.0345 is volume of 1 bag)

Considering 2% Wastage 9.96 =10.87*1.02

Total Cement Bag 9.96

Sand

Sand 1.03 =(3/4)*1.5

Considering 10% Wastage 1.24 =1.125*1.10

Total sand 1.23 Cum

Water

W/C ratio 0.5

Cement 498 9.96*50 (50kg is the weight of one bag of cement

Water 249 =498*0.5

Total water 249 ltr

Environmental engineer

the quantity calculated for 1m3 mortar

the ratio is 1:3 for cement and sand

assume shrinkage factor 1.33

now we can calculate for cement for 1m3 mortar=1*1.33*1/4

=0.3325m3

50kg cement bag volume=50/1440=0.0347m3

so cement weight=0.3325*50kg/0.0347=479.1kg

sand volume=1*1.33*3/4=0.9975cum

now assume water-cement ratio=0.6

so water =0.6*527.01=316.2kg=316.2ltr

Ricardo Kuhn
Construction Foreman

Since you have not mentioned whether the 1:3 cement sand mortar is as per weight batching or volume batching, here I'll describe the process for both the types.

Weight batching:

• Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 and suppose w/c ratio is 0.4
• This means of 1 kg mortar, 0.4 part is water,
• 1 part is cement and 3 parts are sand. 1 part = 1 kg /( 0.4 + 1 + 3)
• This gives, 1 part = 0.227

Quantities of different constituents are,

• water = 0.4 * 1 part =0.4 * 0.227 kg = 0.09 kg
• Cement = 1 part = 0.227 kg
• Sand = 3* 1 part = 0.683 kg

If the sand used for this mix contained adsorbed water equal to 5% of weight of sand, the calculations would change as follows:

• Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 and suppose w/c ratio is 0.4
• This means of 1 kg mortar, 0.4 part is water, 1 part is cement and 3 parts are sand.
• The dry sand weighs only 1/1.05 times ie: .952 times of moist sand. Hence, to maintain 3 parts dry sand, moist sand will have to be added in 3*1.05 part = 3.15 part.
• Of the total 0.4 part water required in the mortar, 5% * 3 part = .15 part will be contributed from adsorbed water in sand and only 0.25 part water will be required to be added externally.
• 1 part = 1 kg /( 0.25 + 1 + 3 + .15)
• This gives, 1 part = 0.227

Quantities of different constituents are,

• water required = 0.25 * 1 part =0.25 * 0.227 kg = 0.056 kg
• Cement = 1 part = 0.227 kg
• Sand = 3.15* 1 part = 0.717 kg

Volume batching:

• Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 by volume and suppose w/c ratio is 0.4
• Assume bulk densities of water as 1 g/c, cement as 1.4 g/cc and fine aggregate as 1.6 g/cc.
• So, mass ratio of cement: sand is 1 part * 1.4 : 3 part * 1.6 .
• This means of 1 kg mortar, 0.4 part is water,
• 1 part is cement and 3.43 parts are sand. 1 part = 1 kg /( 0.4 + 1 + 3.43)
• This gives, 1 part = 0.207

Quantities of different constituents are,

• water = 0.4 * 1 part =0.4 * 0.207 kg = 0.083 kg = 0.83 L of water.
• Cement = 1 part = 0.207 kg = 0.207/ 1.4 = 0.1478 L of cement
• Sand = 3.43* 1 part = 0.71 kg = 0.71/1.6 = 0.444 L of dry sand.

If the sand used for this mix contained adsorbed water equal to 5% of weight of sand, which also caused bulking of sand volume by 20% the calculations would change as follows

• Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 by volume and suppose w/c ratio is 0.4
• Assume bulk densities of water as 1 g/c, cement as 1.4 g/cc and fine aggregate as 1.6 g/cc.
• So, mass ratio of cement: sand is 1 part * 1.4 : 3 part * 1.6 .
• This means of 1 kg mortar, 0.4 part is water, 1 part is cement and 3.43 parts are sand.
• Of the total 0.4 part water required in the mortar, 5% * 3.43 part = .172 part will be contributed from adsorbed water in sand and only 0.228 part water will be required to be added externally. 1 part = 1 kg /( 0.4 + 1 + 3.43)
• This gives, 1 part = 0.207

Quantities of different constituents are,

water = 0.4 * 1 part =0.4 * 0.207 kg = 0.083 kg = 0.83 L of water.

Cement = 1 part = 0.207 kg = 0.207/ 1.4 = 0.1478 L of cement

Changed bulk density of sand is 1.6 * 1.05/1.2 = 1.4 ( because of both adsorbed

Sand = 3.43* 1 part = 0.71 kg = 0.71/1.4 = 0.50 L of moist sand.

Weight of different ingredient of Mortar 1:3 - Here, I will give you a simple way to calculate the weight of different ingredients in a mortar.

The mortar ratio is 1:3, it simply means that the 1 part of cement is present and 3 part of sand is present in that type of mortar. Consider, w/c ratio = 0.4 We are doing all calculation for 1kg of mortar.

1 part = 1/(0.4+1+3) = 0.227 Water Required =0.25 x 0.227 = 0.056kgCement= 0.2247 kgSand =3.15 x0.227 = 0.717 kg Thank You.